Operasi Aljabar Fungsi

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 : 

 

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 : 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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

  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

  33. 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

  34. 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

  35. 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

  36. 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

  37. 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

  38. 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

  39. 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

  40. 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

  41. 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)

  42. 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

  43. 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

  44. 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

  45. 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

  46. 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

  47. 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

  48. 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

  49. 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

  50. 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

  51. 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

  52. 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

  53. 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

  54. 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

  55. 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

  56. 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

  57. 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

  58. 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

  59. 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

  60. 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

  61. 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

  62. 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

  63. 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

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