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Pertanyaan

open parentheses fraction numerator 8 x squared y to the power of negative 4 end exponent over denominator 125 x to the power of negative 1 end exponent y squared end fraction close parentheses to the power of 2 over 3 end exponent equals.... 

  1. 0 comma 16 x squared y to the power of negative 4 end exponent 

  2. 0 comma 16 x y squared 

  3. 1 comma 6 x to the power of negative 2 end exponent y to the power of negative 4 end exponent 

  4. 0 comma 16 x to the power of negative 2 end exponent y to the power of 4 

  5. 1 comma 6 x squared y to the power of negative 4 end exponent 

Pembahasan Soal:

Jika diberikan bilangan riil a dan bilangan rasional p space dan space q maka berlaku sifat berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a to the power of p over a to the power of q end cell equals cell a to the power of p minus q end exponent end cell row cell open parentheses a to the power of p close parentheses to the power of q end cell equals cell a to the power of p q end exponent end cell end table

Sehingga dapat ditentukan hasil dari bentuk berikut. 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 8 x squared y to the power of negative 4 end exponent over denominator 125 x to the power of negative 1 end exponent y squared end fraction close parentheses to the power of 2 over 3 end exponent end cell equals cell open parentheses 8 over 125 x to the power of 2 minus open parentheses negative 1 close parentheses end exponent y to the power of negative 4 minus 2 end exponent close parentheses to the power of 2 over 3 end exponent end cell row blank equals cell open parentheses 2 cubed over 5 cubed x to the power of 2 plus 1 end exponent y to the power of negative 6 end exponent close parentheses to the power of 2 over 3 end exponent end cell row blank equals cell open parentheses open parentheses 2 over 5 close parentheses cubed x cubed y to the power of negative 6 end exponent close parentheses to the power of 2 over 3 end exponent end cell row blank equals cell open parentheses 2 over 5 close parentheses to the power of 3 times 2 over 3 end exponent x to the power of 3 times 2 over 3 end exponent y to the power of negative 6 times 2 over 3 end exponent end cell row blank equals cell open parentheses 0 comma 4 close parentheses squared x squared y to the power of negative 4 end exponent end cell row blank equals cell 0 comma 16 x squared y to the power of negative 4 end exponent end cell end table        

Sehingga diperoleh open parentheses fraction numerator 8 x squared y to the power of negative 4 end exponent over denominator 125 x to the power of negative 1 end exponent y squared end fraction close parentheses to the power of 2 over 3 end exponent equals 0 comma 16 x squared y to the power of negative 4 end exponent.

Jadi, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Afrisno

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui  dan . Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Untuk menyelesaikan soal ini, kalian harus memahami sifat-sifat eksponen berikut:

  • a to the power of p over a to the power of q equals a to the power of p minus q end exponent 
  • open parentheses a to the power of p close parentheses to the power of q equals a to the power of p times q end exponent 

Maka, bentuk sederhananya:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 12 a to the power of negative 5 end exponent b c over denominator 84 a squared b to the power of negative 1 end exponent c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses 12 over 84 open parentheses a to the power of negative 5 minus 2 end exponent close parentheses open parentheses b to the power of 1 minus open parentheses negative 1 close parentheses end exponent close parentheses open parentheses c to the power of 1 minus 1 end exponent close parentheses close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 1 over 7 a to the power of negative 7 end exponent b squared close parentheses to the power of negative 1 end exponent end cell row blank equals cell 7 a to the power of 7 b to the power of negative 2 end exponent end cell row blank equals cell fraction numerator 7 a to the power of 7 over denominator b squared end fraction end cell end table   

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Bentuk sederhana dari (y−3)2×y91​ adalah ....

Pembahasan Soal:

Dengan menggunakan sifat bilangan berpangkat/eksponen:

1. 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

2. a to the power of m over a to the power of n equals a to the power of m minus n end exponent

3. a to the power of negative n end exponent equals 1 over a to the power of n 

Sehingga diperoleh:

(y3)2×y91====y9y6y69y15y151 

Dengan demikian, bentuk sederhana dari (y3)2×y91 adalah y151.

0

Roboguru

Bentuk sederhana dari (2a2b2a3b2​)2 adalah ....

Pembahasan Soal:

Ingat sifat bilangan berpangkat berikut:

anam=am÷an=amn 
(am)n=amn

Maka bentuk bilangan berpangkat di atas dapat disederhanakan seperti berikut, 

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses fraction numerator 2 a cubed b squared over denominator 2 a squared b end fraction close parentheses squared end cell equals cell open parentheses 2 over 2 open parentheses a to the power of 3 minus 2 end exponent close parentheses open parentheses b to the power of 2 minus 1 end exponent close parentheses close parentheses squared end cell row blank equals cell open parentheses a b close parentheses squared end cell row blank equals cell a squared b squared end cell end table end style

Dengan demikian, bentuk sederhana dari (2a2b2a3b2)2 adalah a2b2.

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Sederhanakalanlah dan hitunglah .

Pembahasan Soal:

Dengan menggunakan sifat eksponen aman=anmdan(an)m=anm, maka didapatkan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 to the power of 4 times 9 to the power of negative 2 end exponent times 5 to the power of negative 3 end exponent over denominator 8 times 3 to the power of negative 5 end exponent times 125 to the power of negative 1 end exponent end fraction end cell equals cell fraction numerator 2 to the power of 4 times left parenthesis 3 squared right parenthesis to the power of negative 2 end exponent times 5 to the power of negative 3 end exponent over denominator 2 cubed times 3 to the power of negative 5 end exponent times left parenthesis 5 cubed right parenthesis to the power of negative 1 end exponent end fraction end cell row blank equals cell fraction numerator 2 to the power of 4 times 3 to the power of negative 4 end exponent times 5 to the power of negative 3 end exponent over denominator 2 cubed times 3 to the power of negative 5 end exponent times 5 to the power of negative 3 end exponent end fraction end cell row blank equals cell 2 to the power of 4 minus 3 end exponent times 3 to the power of negative 4 minus left parenthesis negative 5 right parenthesis end exponent times 5 to the power of negative 3 minus left parenthesis negative 3 right parenthesis end exponent end cell row blank equals cell 2 to the power of 1 times 3 to the power of 1 times 5 to the power of 0 end cell row blank equals 6 end table end style

Jadi, hasil dari begin mathsize 14px style fraction numerator 2 to the power of 4 times 9 to the power of negative 2 end exponent times 5 to the power of negative 3 end exponent over denominator 8 times 3 to the power of negative 5 end exponent times 125 to the power of negative 1 end exponent end fraction end style adalah 6.

1

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

Ingat operasi perkalian dan pembagian pada bilangan berpangkat dan juga sifat-sifat dari bilangan berpangkat sendiri.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of 9 cross times 4 to the power of negative 3 end exponent divided by 2 squared end cell equals cell 2 to the power of 9 cross times open parentheses 2 squared close parentheses to the power of negative 3 end exponent divided by 2 squared end cell row blank equals cell 2 to the power of 9 cross times 2 to the power of open parentheses 2 right parenthesis cross times left parenthesis negative 3 close parentheses end exponent divided by 2 squared end cell row blank equals cell 2 to the power of 9 cross times 2 to the power of negative 6 end exponent divided by 2 squared end cell row blank equals cell 2 to the power of 9 plus open parentheses negative 6 close parentheses minus 2 end exponent end cell row blank equals cell 2 to the power of 9 minus 6 minus 2 end exponent end cell row blank equals cell 2 to the power of 1 end cell row blank equals 2 end table

Oleh karena itu, jawaban yang benar adalah B.

2

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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