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

Pertanyaan

Bentuk sederhana dari open parentheses fraction numerator 2 a cubed b squared over denominator 2 a squared b end fraction close parentheses squared adalah ....

  1. begin mathsize 14px style a b end style 

  2. a squared b 

  3. a squared b squared  

  4. a b squared  

  5. begin mathsize 14px style a squared b to the power of 4 end style 

Pembahasan Video:

Pembahasan Soal:

Ingat sifat bilangan berpangkat berikut:

a to the power of m over a to the power of n equals a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent 
open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent

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 open parentheses fraction numerator 2 a cubed b squared over denominator 2 a squared b end fraction close parentheses squared adalah a squared b squared.

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 23 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Hasil dari 9−1:3−2 = . . . .

Pembahasan Soal:

Ingat!

Dalam perpangkatan, berlaku sifat:

  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 space colon space a to the power of n equals a to the power of m minus n end exponent 

Sehingga:

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

Jadi, jawaban yang benar adalah C.

0

Roboguru

Bentuk sederhana dari (ab2c−1a2b4c−2​)3adalah ….

Pembahasan Soal:

Perhatikan 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator a squared b to the power of 4 c to the power of negative 2 end exponent over denominator a b squared c to the power of negative 1 end exponent end fraction close parentheses cubed end cell equals cell fraction numerator a to the power of 6 b to the power of 12 c to the power of negative 6 end exponent over denominator a cubed b to the power of 6 c to the power of negative 3 end exponent end fraction end cell row blank equals cell a cubed b to the power of 6 c to the power of negative 3 end exponent end cell row blank equals cell fraction numerator a cubed b to the power of 6 over denominator c cubed end fraction end cell end table

Jadi, bentuk sederhana dari open parentheses fraction numerator a squared b to the power of 4 c to the power of negative 2 end exponent over denominator a b squared c to the power of negative 1 end exponent end fraction close parentheses cubedadalah fraction numerator a cubed b to the power of 6 over denominator c cubed end fraction.

0

Roboguru

(125x−1y28x2y−4​)32​=....

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.

0

Roboguru

Bentuk sederhana dari bilangan pangkat di bawah ini adalah .... (a−2b−1c2a−1b4c−2​)−3

Pembahasan Soal:

Ingat sifat-sifat bilangan berpangkat berikut ini.

  • a to the power of m over a to the power of n equals a to the power of m minus n end exponent
  • 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
  • a to the power of negative m end exponent equals 1 over a to the power of m

Dengan menerapkan sifat-sifat bilangan berpangkat, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator a to the power of negative 1 end exponent b to the power of 4 c to the power of negative 2 end exponent over denominator a to the power of negative 2 end exponent b to the power of negative 1 end exponent c squared end fraction close parentheses to the power of negative 3 end exponent end cell equals cell open parentheses a to the power of negative 1 minus open parentheses negative 2 close parentheses end exponent space b to the power of 4 minus open parentheses negative 1 close parentheses end exponent space c to the power of negative 2 minus 2 end exponent close parentheses to the power of negative 3 end exponent end cell row blank equals cell open parentheses a to the power of negative 1 plus 2 end exponent space b to the power of 4 plus 1 end exponent space c to the power of negative 2 minus 2 end exponent close parentheses to the power of negative 3 end exponent end cell row blank equals cell open parentheses a to the power of 1 b to the power of 5 c to the power of negative 4 end exponent close parentheses to the power of negative 3 end exponent end cell row blank equals cell a to the power of 1 cross times open parentheses negative 3 close parentheses end exponent space b to the power of 5 cross times open parentheses negative 3 close parentheses end exponent space c to the power of negative 4 cross times open parentheses negative 3 close parentheses end exponent end cell row blank equals cell a to the power of negative 3 end exponent b to the power of negative 15 end exponent c to the power of 12 end cell row blank equals cell fraction numerator c to the power of 12 over denominator a cubed b to the power of 15 end fraction end cell end table 

Jadi, bentuk sederhana dari bilangan berpangkat di atas adalah fraction numerator c to the power of 12 over denominator a cubed b to the power of 15 end fraction.

0

Roboguru

Sederhanakan operasi bilangan berpangkat berikut! a. (a43​×a−83​÷a61​)8 b. (a21​×a−32​÷b61​)−6

Pembahasan Soal:

Ingatlah sifat-sifat bilangan berpangkat berikut.

left parenthesis straight i right parenthesis space 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 left parenthesis ii right parenthesis space a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent left parenthesis iii right parenthesis space open parentheses a divided by b close parentheses to the power of m equals a to the power of m divided by b to the power of m left parenthesis iv right parenthesis space 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  

Berdasarkan sifat-sifat di atas, maka:

a)

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 3 over 4 end exponent cross times a to the power of negative 3 over 8 end exponent divided by a to the power of 1 over 6 end exponent close parentheses to the power of 8 end cell equals cell open parentheses a to the power of 3 over 4 plus open parentheses negative 3 over 8 close parentheses minus 1 over 6 end exponent close parentheses to the power of 8 end cell row blank equals cell open parentheses a to the power of fraction numerator 18 minus 9 minus 4 over denominator 24 end fraction end exponent close parentheses to the power of 8 end cell row blank equals cell open parentheses a to the power of 5 over 24 end exponent close parentheses to the power of 8 end cell row blank equals cell a to the power of 5 over 24 cross times 8 end exponent end cell row blank equals cell a to the power of 40 over 24 end exponent end cell row blank equals cell a to the power of 5 over 3 end exponent end cell end table

b)

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 1 half end exponent cross times a to the power of negative 2 over 3 end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell equals cell open parentheses a to the power of 1 half plus open parentheses negative 2 over 3 close parentheses end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell open parentheses a to the power of fraction numerator 3 minus 4 over denominator 6 end fraction end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell open parentheses a to the power of negative 1 over 6 end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell open parentheses a to the power of negative 1 over 6 end exponent close parentheses to the power of negative 6 end exponent divided by open parentheses b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell a to the power of negative fraction numerator 1 over denominator up diagonal strike 6 end fraction cross times open parentheses negative up diagonal strike 6 close parentheses end exponent divided by b to the power of fraction numerator 1 over denominator up diagonal strike 6 end fraction cross times open parentheses negative up diagonal strike 6 close parentheses end exponent end cell row blank equals cell a divided by b to the power of negative 1 end exponent end cell row blank equals cell a divided by 1 over b end cell row blank equals cell a b end cell end table

Dengan demikian, bentuk sederhana dari  open parentheses a to the power of 3 over 4 end exponent cross times a to the power of negative 3 over 8 end exponent divided by a to the power of 1 over 6 end exponent close parentheses to the power of 8 equals a to the power of 5 over 3 end exponent, dan open parentheses a to the power of 1 half end exponent cross times a to the power of negative 2 over 3 end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent equals a b.

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