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Sederhanakan g. 8k5:2k2=...

Pertanyaan

Sederhanakan

g. 8k5:2k2=...

Pembahasan Soal:

Ingat!

Sifat bilangan berpangkat:

  • am:an=amn 

Sehingga:

8k5:2k2===2k28k54k524k3 

Dengan demikian, bentuk sederhana dari 8k5:2k2 adalah 4k3.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 30 Agustus 2021

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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

Diketahui  dan . Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Ingat sifat-sifat bilangan berpangkat berikut ini:

left parenthesis straight i right parenthesis space a to the power of m over a to the power of n equals a to the power of m minus n end exponent left parenthesis ii right parenthesis space a to the power of m a to the power of n equals a to the power of m plus n end exponent 

Bedarakan sifat-sifat di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 27 to the power of negative 1 end exponent a to the power of negative 3 end exponent b to the power of negative 5 end exponent over denominator begin display style 1 third end style a squared b to the power of 4 end fraction close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses fraction numerator open parentheses 3 cubed close parentheses to the power of negative 1 end exponent a to the power of negative 3 end exponent b to the power of negative 5 end exponent over denominator 3 to the power of negative 1 end exponent a squared b to the power of 4 end fraction close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 3 to the power of negative 3 minus open parentheses negative 1 close parentheses end exponent a to the power of negative 3 minus 2 end exponent b to the power of negative 5 minus 4 end exponent close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 3 to the power of negative 2 end exponent a to the power of negative 5 end exponent b to the power of negative 9 end exponent close parentheses to the power of negative 1 end exponent end cell row blank equals cell 3 squared a to the power of 5 b to the power of 9 end cell row blank equals cell 9 a to the power of 5 b to the power of 9. end cell end table 

Jadi, bentuk sederhana dari open parentheses fraction numerator 27 to the power of negative 1 end exponent a to the power of negative 3 end exponent b to the power of negative 5 end exponent over denominator begin display style 1 third end style a squared b to the power of 4 end fraction close parentheses to the power of negative 1 end exponent adalah 9 a to the power of 5 b to the power of 9.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Sederhanakan bentuk-bentuk berikut! 8a−3b5a−6b8​

Pembahasan Soal:

Ingat kembali sifat bilangan berpangkat berikut.

am÷anam==amnam1,a=0

Sehingga, diperoleh perhitungan:

8a3b5a6b8===8a6(3)b858a3b38a3b3

Jadi, bentuk sederhana dari 8a3b5a6b8 adalah8a3b3.

0

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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