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Bentuk sederhana dari (3​−2)(3​+5)adalah ....

Pertanyaan

Bentuk sederhana dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spaceadalah ....

Pembahasan Video:

Pembahasan Soal:

Untuk menemukan bentuk sederhana dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spaceyaitu dengan mengalikan secara distributif dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spacesehingga diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses space end cell equals cell open parentheses square root of 3 cross times square root of 3 close parentheses plus open parentheses square root of 3 cross times 5 close parentheses plus open parentheses negative 2 cross times square root of 3 close parentheses plus open parentheses negative 2 cross times 5 close parentheses end cell row cell open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses end cell equals cell 3 plus 5 square root of 3 minus 2 square root of 3 minus 10 end cell row cell open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses end cell equals cell 3 square root of 3 minus 7 end cell end table end style


Jadi bentuk sederhana dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spaceadalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 3 square root of 3 minus 7 end cell end table

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan hasil perkalian akar-akar bilangan berikut! e. (24a5​+b​)(24a5​−b​)

Pembahasan Soal:

Ingat kembali rumus berikut.

  • open parentheses a to the power of m close parentheses to the power of n equals a to the power of m cross times n end exponent 
  • n-th root of a to the power of m end root equals a to the power of m over n end exponent 
  • a to the power of m cross times a to the power of n equals a to the power of m plus n end exponent 
  • square root of a squared cross times b end root equals a square root of b
  • open parentheses square root of a minus square root of b close parentheses open parentheses square root of a plus square root of b close parentheses equals open parentheses square root of a close parentheses squared minus open parentheses square root of b close parentheses squared 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 fourth root of a to the power of 5 end root plus square root of b close parentheses open parentheses 2 fourth root of a to the power of 5 end root minus square root of b close parentheses end cell equals cell 2 squared open parentheses fourth root of a to the power of 5 end root close parentheses squared minus open parentheses square root of b close parentheses squared end cell row blank equals cell 4 open parentheses a to the power of 5 over 4 end exponent close parentheses squared minus b end cell row blank equals cell 4 open parentheses a to the power of 5 over 4 cross times 2 end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a to the power of 5 over 2 end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a to the power of 2 1 half end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a to the power of 2 plus 1 half end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a squared cross times a to the power of 1 half end exponent close parentheses minus b end cell row blank equals cell 4 a squared square root of a minus b end cell end table 

Jadi, hasil dari open parentheses 2 fourth root of a to the power of 5 end root plus square root of b close parentheses open parentheses 2 fourth root of a to the power of 5 end root minus square root of b close parentheses adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table squared table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative b end cell end table.

0

Roboguru

Hasil perkalian (6​+3​)(23​−6​) adalah ....

Pembahasan Soal:

Berdasarkan sifat distributif perkalian yaitu 

open parentheses a plus b close parentheses open parentheses c plus d close parentheses equals a b plus a d plus b c plus b d

Maka open parentheses square root of 6 plus square root of 3 close parentheses open parentheses 2 square root of 3 minus square root of 6 close parentheses diperoleh sebagai berikut.

begin mathsize 12px style open parentheses square root of 6 plus square root of 3 close parentheses open parentheses 2 square root of 3 minus square root of 6 close parentheses equals 2 square root of 3 cross times square root of 6 minus square root of 6 cross times square root of 6 plus square root of 3 cross times 2 square root of 3 minus square root of 3 cross times square root of 6 end style

Berdasarakan aturan perkalian bentuk akar dengan akar pangkat sama yaitu

n-th root of a cross times n-th root of b equals n-th root of a cross times b end root comma space a not equal to 0 space text dan end text space b not equal to 0

Sehingga 

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 6 plus square root of 3 close parentheses open parentheses 2 square root of 3 minus square root of 6 close parentheses end cell equals cell 2 square root of 3 cross times square root of 6 minus square root of 6 cross times square root of 6 plus square root of 3 cross times 2 square root of 3 minus square root of 3 cross times square root of 6 end cell row blank equals cell 2 square root of 3 cross times 6 end root minus square root of 6 cross times 6 end root plus 2 square root of 3 cross times 3 end root minus square root of 3 cross times 6 end root end cell row blank equals cell 2 square root of 18 minus square root of 36 plus 2 square root of 9 minus square root of 18 end cell row blank equals cell 2 square root of 18 minus square root of 18 minus 6 plus 2 cross times 3 end cell row blank equals cell open parentheses 2 minus 1 close parentheses square root of 18 up diagonal strike negative 6 plus 6 end strike end cell row blank equals cell square root of 18 end cell end table end style

Jika n-th root of a dapat difaktorkan menjadi n-th root of p cross times q end root dengan p space text atau end text space q bilangan berpangkat n maka n-th root of a dapat disederhanakan sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 6 plus square root of 3 close parentheses open parentheses 2 square root of 3 minus square root of 6 close parentheses end cell equals cell square root of 18 end cell row blank equals cell square root of 9 cross times 2 end root end cell row blank equals cell square root of 3 squared cross times 2 end root end cell row blank equals cell 3 square root of 2 end cell end table

Jadi Hasil perkalian open parentheses square root of 6 plus square root of 3 close parentheses open parentheses 2 square root of 3 minus square root of 6 close parentheses adalah 3 square root of 2.

Oleh karena itu, jawaban yang benar adalah B.

1

Roboguru

Tentukan hasil perkalian akar-akar bilangan berikut! c. (25​−2​)(25​+2​)

Pembahasan Soal:

Ingat kembali rumus berikut.

open parentheses square root of a minus square root of b close parentheses open parentheses square root of a plus square root of b close parentheses equals open parentheses square root of a close parentheses squared minus open parentheses square root of b close parentheses squared 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 square root of 5 minus square root of 2 close parentheses open parentheses 2 square root of 5 plus square root of 2 close parentheses end cell equals cell 2 squared open parentheses square root of 5 close parentheses squared minus open parentheses square root of 2 close parentheses squared end cell row blank equals cell 4 cross times 5 minus 2 end cell row blank equals cell 20 minus 2 end cell row blank equals 18 end table 

Jadi, hasil dari open parentheses 2 square root of 5 minus square root of 2 close parentheses open parentheses 2 square root of 5 plus square root of 2 close parentheses adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 18 end table.

0

Roboguru

Jika x=1+3​ dan y=1−3​  maka nilai xy adalah...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell x y end cell equals cell left parenthesis 1 plus square root of 3 right parenthesis left parenthesis 1 minus square root of 3 right parenthesis end cell row blank equals cell 1 minus square root of 3 plus square root of 3 minus 3 end cell row blank equals cell 1 minus 3 end cell row blank equals cell negative 2 end cell end table  

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Tentukan hasil perkalian akar-akar bilangan berikut! a. (12​+7)(12​−7)

Pembahasan Soal:

Ingat kembali rumus berikut.

open parentheses square root of a plus b close parentheses open parentheses square root of a minus b close parentheses equals open parentheses square root of a close parentheses squared minus b squared 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 12 plus 7 close parentheses open parentheses square root of 12 minus 7 close parentheses end cell equals cell open parentheses square root of 12 close parentheses squared minus 7 squared end cell row blank equals cell 12 minus 49 end cell row blank equals cell negative 37 end cell end table 

Jadi, hasil dari open parentheses square root of 12 plus 7 close parentheses open parentheses square root of 12 minus 7 close parentheses adalah negative 37.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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