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Sederhanakan k. (−3k3)3=...

Pertanyaan

Sederhanakan

k. (3k3)3=...

Pembahasan Soal:

Ingat!

Sifat bilangan berangkat:

  • (a×b)n=an×bn 
  • (am)n=am×n   

Sehingga:

(3k3)3===(3)3×k3×327×k927k9 

Dengan demikian, bentuk sederhana dari (3k3)3 adalah 27k9.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Jika  dan , nilai  adalah ....

Pembahasan Soal:

Sifat bilangan berpangkatan yang digunakan:
open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent 

Dengan menerapkan sifat pemangkatan bilangan berpangkat, diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 5 to the power of 10 x end exponent end cell equals 1600 row cell 5 to the power of 2 times 5 x end exponent end cell equals 1600 row cell open parentheses 5 to the power of 5 x end exponent close parentheses squared end cell equals 1600 row cell square root of open parentheses 5 to the power of 5 x end exponent close parentheses squared end root end cell equals cell square root of 1600 end cell row cell 5 to the power of 5 x end exponent end cell equals 40 end table end style 

Selanjutnya, dengan menerapkan sifat-sifat bilangan berpangkat, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 5 to the power of x minus 1 end exponent close parentheses to the power of 5 over 8 to the power of open parentheses negative square root of y close parentheses end exponent end cell equals cell 5 to the power of 5 open parentheses x minus 1 close parentheses end exponent over open parentheses 2 cubed close parentheses to the power of negative square root of y end exponent end cell row blank equals cell 5 to the power of 5 x minus 5 end exponent over 2 to the power of negative 3 square root of y end exponent end cell row blank equals cell fraction numerator begin display style 5 to the power of 5 x end exponent over 5 to the power of 5 end style over denominator begin display style 1 over 2 to the power of 3 square root of y end exponent end style end fraction end cell row blank equals cell 5 to the power of 5 x end exponent over 5 to the power of 5 cross times 2 to the power of 3 square root of y end exponent end cell row blank equals cell 5 to the power of 5 x end exponent over 5 to the power of 5 cross times open parentheses 2 to the power of square root of y end exponent close parentheses cubed end cell row blank equals cell 40 over 5 to the power of 5 cross times 25 cubed end cell row blank equals cell 40 over 5 to the power of 5 cross times open parentheses 5 squared close parentheses cubed end cell row blank equals cell 40 over 5 to the power of 5 cross times 5 to the power of 2 times 3 end exponent end cell row blank equals cell 40 over 5 to the power of 5 cross times 5 to the power of 6 end cell row blank equals cell 40 cross times 5 to the power of 6 minus 5 end exponent end cell row blank equals cell 40 cross times 5 to the power of 1 end cell row blank equals cell 40 cross times 5 end cell row blank equals 200 end table end style 

Jadi, nilai begin mathsize 14px style open parentheses 5 to the power of x minus 1 end exponent close parentheses to the power of 5 over 8 to the power of open parentheses negative square root of y close parentheses end exponent end style adalah 200.

0

Roboguru

Bentuk sederhana dari adalah...

Pembahasan Soal:

Gunakan sifat eksponen,

1. space open parentheses straight a to the power of straight n close parentheses to the power of straight m equals straight a to the power of nm 2. space space straight a to the power of straight n over straight a to the power of straight m equals straight a to the power of straight n minus straight m end exponent

Dari sifat-sifat diatas, bentuk begin mathsize 14px style open parentheses x squared y cubed z to the power of 4 close parentheses cubed over open parentheses x to the power of negative 2 end exponent y to the power of 4 z cubed close parentheses squared end style dapat disederhanakan menjadi

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x squared y cubed z to the power of 4 close parentheses cubed over open parentheses x to the power of negative 2 end exponent y to the power of 4 z cubed close parentheses squared end cell equals cell fraction numerator x to the power of 6 y to the power of 9 z to the power of 12 over denominator x to the power of negative 4 end exponent y to the power of 8 z to the power of 6 end fraction end cell row blank equals cell x to the power of 6 minus left parenthesis negative 4 right parenthesis end exponent y to the power of 9 minus 8 end exponent z to the power of 12 minus 6 end exponent end cell row blank equals cell x to the power of 10 y z to the power of 6 end cell end table end style

Jadi, bentuk sederhana dari begin mathsize 14px style open parentheses x squared y cubed z to the power of 4 close parentheses cubed over open parentheses x to the power of negative 2 end exponent y to the power of 4 z cubed close parentheses squared end style  adalah Error converting from MathML to accessible text.

1

Roboguru

Tentukan bentuk sederhana dan positif

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

a to the power of n equals stack a cross times a cross times a cross times horizontal ellipsis cross times a with underbrace below table row blank cell space space space space space space space space end cell cell n space text faktor end text end cell end table

Ingat sifat bilangan berpangkat berikut.

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 b close parentheses to the power of m equals a to the power of m times b to the power of m

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 comma space b not equal to 0

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent

a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0

sehingga penyelesaian soal di atas, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open square brackets fraction numerator 7 x cubed y to the power of negative 4 end exponent z to the power of negative 6 end exponent over denominator 84 x to the power of negative 7 end exponent y to the power of negative 1 end exponent z to the power of negative 4 end exponent end fraction close square brackets squared end cell row blank equals cell open square brackets 1 over 12 times x to the power of 3 minus open parentheses negative 7 close parentheses end exponent times y to the power of negative 4 minus open parentheses negative 1 close parentheses end exponent times z to the power of negative 6 minus open parentheses negative 4 close parentheses end exponent close square brackets squared end cell row blank equals cell open square brackets 1 over 12 times x to the power of 10 times y to the power of negative 3 end exponent times z to the power of negative 2 end exponent close square brackets squared end cell row blank equals cell open parentheses 1 over 12 close parentheses squared open parentheses x to the power of 10 close parentheses squared open parentheses y to the power of negative 3 end exponent close parentheses squared open parentheses z to the power of negative 2 end exponent close parentheses squared end cell row blank equals cell 1 over 144 x to the power of 20 y to the power of negative 6 end exponent z to the power of negative 4 end exponent end cell row blank equals cell fraction numerator x to the power of 20 over denominator 144 y to the power of 6 z to the power of 4 end fraction end cell end table

Dengan demikian, bentuk sederhana dari table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets fraction numerator 7 x cubed y to the power of negative 4 end exponent z to the power of negative 6 end exponent over denominator 84 x to the power of negative 7 end exponent y to the power of negative 1 end exponent z to the power of negative 4 end exponent end fraction close square brackets squared end cell equals cell fraction numerator x to the power of 20 over denominator 144 y to the power of 6 z to the power of 4 end fraction end cell end table

0

Roboguru

Bentuk sederhana dari  adalah ...

Pembahasan Soal:

Mencari bentuk sederhana dari open parentheses fraction numerator 12 a cubed b squared over denominator 4 a b to the power of 5 end fraction close parentheses to the power of negative 1 end exponent:

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

Jadi, bentuk sederhana dari open parentheses fraction numerator 12 a cubed b squared over denominator 4 a b to the power of 5 end fraction close parentheses to the power of negative 1 end exponent adalah fraction numerator b cubed over denominator 3 a squared end fraction.

Dengan demikian, jawaban yang tepat adalah A.

0

Roboguru

Bentuk sederhana dari adalah...

Pembahasan Soal:

begin mathsize 14px style fraction numerator a to the power of negative 5 end exponent b to the power of negative 1 end exponent c to the power of negative 4 end exponent over denominator open parentheses a b c close parentheses to the power of negative 6 end exponent end fraction equals fraction numerator a to the power of negative 5 end exponent b to the power of negative 1 end exponent c to the power of negative 4 end exponent over denominator a to the power of negative 6 end exponent b to the power of negative 6 end exponent c to the power of negative 6 end exponent end fraction space space space space space space space space space space space space space space space space space equals a to the power of negative 5 minus open parentheses negative 6 close parentheses end exponent b to the power of negative 1 minus open parentheses negative 6 close parentheses end exponent c to the power of negative 4 minus open parentheses negative 6 close parentheses end exponent space space space space space space space space space space space space space space space space space equals a b to the power of 5 c squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end style 

Jadi, jawaban yang tepat adalah C

2

Roboguru

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