Operasi Aljabar Fungsi

13 Januari 2018 Ika Desi B Fungsi 63

Operasi-Aljabar-Fungsi

Hai untuk kelas online kali ini kita akan belajar tentang operasi aljabar fungsi. Silakan simak video di bawah ini. Jika ada yang kurang jelas silakan tanyakan di bagian komentar ya.

Video : 

[embedyt] https://www.youtube.com/watch?v=AF8P3mqmJU0[/embedyt]

 

Diskusi :

Diketahui f(x)= x + 2 dan g(x)= x^{2} - 4

Tentukanlah :

  1. (f + g) (1)
  2. (f - g) (2)
  3. (f . g) (3)
  4. (f / g) (4)

Tulis jawabanmu di kolom komentar di bawah. Ingat tulis juga nama, kelas, dan nomor absenmu ya.

 

—— Selamat Belajar ——

 

63 Komentar

  1. Nama : I KD GD AGUS WIDIA GUNA PUTRA
    No : 08/09
    Kelas : X IPS 4

    Operasi Aljabar Fungsi :
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  2. Nama : I Made Krisna Kusuma
    No : 11/12
    Kelas : X IPS 4

    Operasi Aljabar Fungsi :
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  3. Nama: Ni luh rechita yuniartini
    No: 25/24
    Kelas:X ips 4

    Operasi aljabar fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1.(f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0

    2.(f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4

    3.(f.3)(3)=(x+2)(x^2-4)=x^3-4+2x^2-8=y^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25

    4(f/g)(4)=(x+2)(x^2-4)=(x+2)/(x-2)(x+2)=1/(x-2)=1/(4-2)=1/2

    Balas

  4. Nama : I Made Erik Kusuma Budarta
    Kelas : X IPS 4
    No:11
    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  5. Nama: Ni Luh Ayu Aprilia Yudiartini
    No: 22/21
    Kelas: X IPS 4

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  6. Nama: Ni komang triana Dewi
    No. : 21
    Kls. : XIPS4
    * OPERASI ALJABAR FUNGSI
    Diketahui
    f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  7. Nama : Ni Luh Bintang Rahayu Anggrelita
    No : 23
    Kelas : X IPS 4

    *Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  8. NAMA ; NI MADE AYU PUSPA WIDYANINGSIH
    NO : 26/25
    KELAS: X IPS 4

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  9. Nama : Anak Agung Dwi Putri Anjani
    No : 01
    Kelas : X IPS 4

    OPERASI ALJABAR FUNGSI

    Diketahui = f(x)=x+2
    = g(x)=x²-4
    Jawaban:

    1. (f+g)(1) = x+2+x²-4
    = 1+2+1²-4
    = 3+1-4
    = 0

    2. (f-g)(2) = x+2-(x²-4)
    = 2+2-(2²-4)
    = 2+2-4+4
    = 4-8
    = -4

    3. (f.g)(3) = (x+2)(x²-4)
    = (3+2)(3²-4)
    = 3+2.9-4
    = 5.5
    = 25

    4. (f/g)(4) = (x+2)/(x²-4)
    = (4+2)/(4²-4)
    = 6 / 16-4
    = 6/12
    = 1/2

    Balas

  10. NAMA; NI WAYAN SHANTI SAFITRI
    NO; 27/28
    KELAS; X IPS 4

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  11. Nama : Ni Komang Nicken Eliana Puspita Dewi
    No : 23
    Kelas : X IIS 6

    Operasi Aljabar Fungsi :
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  12. Nama : made bagus oka wiguna
    No : 15
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  13. Nama : Dewa Gede Dika Mahardana
    No :01
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  14. Nama : I Made Pande Agus Sudarma Putra Yoga
    No : 07
    Kls : X IPS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  15. Nama : I Made Mereta Adnyana Wibawa
    No : 06
    Kelas : X IPS 6
    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  16. Nama : I made rai prayoga
    No : 11
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  17. Nama : I Made Panji Adnyana
    Kelas : X IIS 6
    No : 09
    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  18. Nama : Made Arista Intan Cahyani
    Kelas : X IIS 6
    No : 16
    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  19. Nama : I NYOMAN MERTA DANA
    No : 11
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  20. Nama : I Putu Gede Indra Budiarsana
    No :14
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  21. Nama : Ni luh putu ari astiarin
    No :25
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  22. Nama : Ni Kadek Hani Dwi Rian
    No :21
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  23. Nama:Kadek Yogi Arleana Putra
    No:15
    Kelas: X Ips 5

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)

    Balas

  24. nama : i made yudi prasetya
    kelas : x ips 6
    no : 12

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  25. nama : i kadek bagus mahardika
    kelas : x ips 5
    no : 7

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  26. OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  27. NAMA : I GUSTI BAGUS HARI MERTA WIJAYA
    KELAS : X IPS 5
    NO : 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  28. Putu adi pratama
    28
    X ipa 5

    Operasi aljabar fungsii
    1.f(x)=x+2 dan g(x)=x^2-4=2=0
    2. -x^2+x+6=-2^+2+6=-4+2+6=4
    3. 27+18-12-8
    4. 1/2

    Balas

  29. NAMA : I GEDE YUDHA ADNYANA
    NO : 05
    KELAS : X IPS 5

    Operasi Aljabar Fungsi

    Diketahui f(x)=x+2 dan g(x)=x^2-4

    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  30. NAMA : PUTU WIDIANA
    KELAS : X IPS 5
    NO : 30

    Operasi Aljabar Fungsi

    Diketahui f(x)=x+2 dan g(x)=x^2-4

    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  31. NAMA : DESAK MADE MITA DWIPAYANI
    NO. ABSEN : 02
    KELAS : X IPS 5

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  32. Nama : Febiyana Agustina
    No : 02
    Kelas : X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  33. nama: putu pande lisna maharani
    absen: 29
    kelas : X IPS5

    dik : f (x) = x + 2
    g(x) = x^ – 4

    jawaban :
    1. (f + g ) (1) =f (1) + g (1) = (1+2) + (1^2 – 4) = 3+(-3) =0
    2. (f – g ) (2) = f(2) – g (2) = (2+2) – (2^2 – 4 ) = 4- (4-4) =4-0 =4
    3. (f.g) (3) = f(3) . g(3) = (3+2) . (3^2 – 4) = (3+2) . (9-4) = (5) . (5) = 25
    4. (f/g) (4) = f (4) / g (4) =( 4 + 2 ) / (4^2-4) = 6/12 = 1/2

    Balas

  34. Nama:I Nyoman Awang Aditya
    No:12
    Kls:x ips 5
    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2= (1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  35. Nama :I kadek suryadana
    No :06
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  36. Nama : Ni Kadek Dwi Handayani
    No :20
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  37. Nama : i putu ananta wijaya
    Kelas : X IIS 6
    Absen : 12

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  38. nama : i nyoman awang aditya
    n0 : 12
    kelas : x ips 5

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  39. nama : ni kadek putri widnyani
    no : 19
    kelas : x ips 5

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  40. Nama : I Putu Agus Krisna Jayantara
    No :13
    Kls : X Ips 5
    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  41. Nama : niluhnoviarisuciptadewi
    No: 21
    Kls : x ips 5

    dik : f (x) = x + 2
    g(x) = x^ – 4

    1. (f + g ) (1) =f (1) + g (1) = (1+2) + (1^2 – 4) =
    3+(-3) =0
    2. (f – g ) (2) = f(2) – g (2) = (2+2) – (2^2 – 4 ) =
    4- (4-4) =4-0 =4
    3. (f.g) (3) = f(3) . g(3) = (3+2) . (3^2 – 4) = (3+2)
    .(9-4) = (5) . (5) = 25
    4. (f/g) (4) = f (4) / g (4) = 1/2

    Balas

  42. nama : clara ni kadek ririn julitayanti
    no : 01
    kelas : x ips 5

    dik : f (x) = x + 2
    g(x) = x^ – 4

    jawaban :
    1. (f + g ) (1) =f (1) + g (1) = (1+2) + (1^2 – 4) = 3+(-3) =0
    2. (f – g ) (2) = f(2) – g (2) = (2+2) – (2^2 – 4 ) = 4- (4-4) =4-0 =4
    3. (f.g) (3) = f(3) . g(3) = (3+2) . (3^2 – 4) = (3+2) . (9-4) = (5) . (5) = 25
    4. (f/g) (4) = f (4) / g (4) =( 4 + 2 ) / (4^2-4) = 6/12 = 1/2

    Balas

  43. Nama : Ni Nyoman Ayu Trisnawati
    No. : 24
    Kelas : X ips 5

    OPERASI ALJABAR FUNGSI

    Diketahui :
    f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  44. Nama : Ni Putu Amanda Aprelia
    No : 25
    Kelas : X IPS 5

    OPERASI ALJABAR FUNGSI

    Diketahui :
    f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2= (1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  45. Nama : Ni Kadek Purwasih
    No : 18
    Kelas : X ips 5

    OPERASI ALJABAR FUNGSI

    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2= (1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  46. Nama:i made rai andi permana
    No:10
    Kls:x ips 5
    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2= (1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  47. Nama:I Gede Indra Palguna
    No:04
    Kls:x ips 5
    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2= (1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  48. NAMA : KADEK PRADNYANDIKA WILLYAN BASTARA
    NOMOR : 09
    KELAS : X IPS 5

    Operasi Aljabar Fungsi

    Diketahui f(x)=x+2 dan g(x)=x^2-4

    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  49. Nama : Ni Kadek Kristina Dewi
    No : 17
    Kelas : X IPS 5

    Operasi Aljabar Fungsi

    Diketahui f(x)=x+2 dan g(x)=x^2-4

    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  50. NOVIA DWI ANDINI

    NO ABSEN: 27

    KELAS : X IPS 5

    Operasi aljabar fungsii
    1.f(x)=x+2 dan g(x)=x^2-4=2=0
    2. -x^2+x+6=-2^+2+6=-4+2+6=4
    3. 27+18-12-8=25
    5. 1/2

    Balas

  51. Nama : Ni Komang Ayu Trisiadewi
    No : 20
    Kelas : X IPS 5

    Operasi Aljabar Fungsi

    Diketahui f(x)=x+2 dan g(x)=x^2-4

    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  52. Nama : Ni Made Novi Yunita
    No : 23
    Kelas : X ips 5

    operasi aljabar fungsi
    diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  53. Nama : I Putu Diva Ari Prabawa
    No : 14
    Kelas : X IPS 5

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2= (1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  54. NI PUTU MIRAH PATMADEWI
    Operasi aljabar fungsii
    1.f(x)=x+2 dan g(x)=x^2-4=2=0
    2. -x^2+x+6=-2^+2+6=-4+2+6=4
    3. 27+18-12-8
    4. 1/2

    Balas

  55. Nama: ni putu aninda putri
    No: 28
    Kelas: X iis 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  56. Nama: ni made ayu sriwidari
    No: 26
    Kelas: X iis 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  57. Nama : ni putu kristanti eka putri
    Kelas : x iis 6
    Absen : 29

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  58. Nama: ni kadek ayu diah arisanti
    No: 19
    Kelas: X iis 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  59. Nama : Ni Kadek Kusuma Dewi
    No : 22
    Kelas : X IIS 6

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  60. Nama : Ni Made Dita Pradnyani
    No : 27
    Kelas : X IIS 6

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  61. Nama : Ni Luh Bintang Maha Putri
    No : 24
    Kelas : X IIS 6

    Operasi Aljabar Fungsi
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  62. Nama : Ni Luh Bintang Maha Putri
    No : 24
    Kelas : X IIS 6

    Operasi Aljabar Fungsi :
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

  63. Nama : Ni Putu Windayanti
    No :30
    Kelas :X IIS 6

    OPERASI ALJABAR FUNGSI
    Diketahui f(x)=x+2 dan g(x)=x^2-4
    1. (f+g)(1)=x+2+x^2-4=x^2+x-2=(1)^2+1-2=1+1-2=0
    2. (f-g)(2)=x+2-(x^2-4)=x+2-x^2+4=-x^2+x+6=-(2)^2+2+6=-4+2+6=4
    3. (f.g)(3)=(x+2)(x^2-4)=x^3-4x+2x^2-8=x^3+2x^2-4x-8=(3)^3+2(3)^2-4(3)-8=27+18-12-8=25
    4. (f/g)(4)=(x+2)/(x^2-4)=(x+2)/((x-2)(x+2))=1/(x-2)=1/(4-2)=1/2

    Balas

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