Roboguru

Sederhanakan f. x10:x3=...

Pertanyaan

Sederhanakan

f. x10:x3=...

Pembahasan Soal:

Ingat!

Sifat bilangan berpangkat:

  • am:an=amn 

Sehingga:

x10:x3==x103x7 

Dengan demikian, bentuk sederhana dari x10:x3 adalah x7.

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

Gunakan sifat-sifat operasi bilangan berpangkat untuk menyederhanakan operasi aljabar berikut. a.

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 b to the power of m

sehingga penyelesaian soal di atas, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 5 n close parentheses to the power of 5 divided by 5 squared n to the power of 4 end cell equals cell 5 to the power of 5 n to the power of 5 divided by 5 squared n to the power of 4 end cell row blank equals cell 5 to the power of 5 minus 2 end exponent n to the power of 5 minus 4 end exponent end cell row blank equals cell 5 cubed n end cell end table

Dengan demikian, bentuk sederhana dari table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 5 n close parentheses to the power of 5 divided by 5 squared n to the power of 4 end cell equals cell 5 cubed n end cell end table 

0

Roboguru

Bentuk sederhana dari adalah....

Pembahasan Soal:

Ingat kembali:

konsep sifat eksponen:

a to the power of n equals stack stack a cross times a cross times a cross times... cross times a with underbrace below with n below 1 over a to the power of n equals a to the power of negative n end exponent open parentheses a to the power of n close parentheses to the power of m equals a to the power of n times m end exponent a to the power of m over a to the power of n equals a to the power of m minus n end exponent m-th root of a to the power of n end root equals a to the power of n over m end exponent

Sehingga, diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator square root of 81 a to the power of negative 12 end exponent b to the power of 10 c to the power of 5 end root over denominator cube root of negative 27 a to the power of negative 8 end exponent b to the power of 6 end root end fraction end cell equals cell open parentheses 3 to the power of 4 a to the power of negative 12 end exponent b to the power of 10 c to the power of 5 close parentheses to the power of begin display style 1 half end style end exponent over open parentheses open parentheses negative 3 close parentheses cubed a to the power of negative 8 end exponent b to the power of 6 close parentheses to the power of begin display style 1 third end style end exponent end cell row blank equals cell fraction numerator 3 squared a to the power of negative 6 end exponent b to the power of 5 c to the power of begin display style 5 over 2 end style end exponent over denominator open parentheses negative 3 close parentheses a to the power of begin display style fraction numerator negative 8 over denominator 3 end fraction end style end exponent b squared end fraction end cell row blank equals cell negative 3 to the power of 2 minus 1 end exponent a to the power of negative 6 minus fraction numerator negative 8 over denominator 3 end fraction end exponent b to the power of 5 minus 2 end exponent c to the power of 5 over 2 end exponent end cell row blank equals cell negative 3 a to the power of negative 10 over 3 end exponent b cubed c to the power of 5 over 2 end exponent end cell row blank equals cell negative fraction numerator 3 b cubed c to the power of 5 over 2 end exponent over denominator a to the power of 10 over 3 end exponent end fraction end cell end table

Dengan demikian, bentuk sederhana dari bentuk akar tersbut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 b cubed c to the power of 5 over 2 end exponent over denominator a to the power of 10 over 3 end exponent end fraction end cell end table

Jadi, jawaban yang tepat adalah C

0

Roboguru

Bentuk sederhana dari  adalah ...

Pembahasan Soal:

Gunakan sifat-sifat operasi hitung bilangan berpangkat untuk mengerjakan soal tersebut.


table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator a to the power of negative 2 end exponent times b to the power of 8 times c to the power of 10 over denominator a to the power of 12 times b to the power of negative 4 end exponent times c to the power of 6 end fraction close parentheses end cell equals cell open parentheses a to the power of negative 2 minus 12 end exponent times b to the power of 8 minus left parenthesis negative 4 right parenthesis end exponent times c to the power of 10 minus 6 end exponent close parentheses end cell row blank equals cell open parentheses a to the power of negative 14 end exponent times b to the power of 12 times c to the power of 4 close parentheses end cell row blank equals cell open parentheses fraction numerator b to the power of 12 times c to the power of 4 over denominator a to the power of 14 end fraction close parentheses end cell end table


Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

....

Pembahasan Soal:

Ingat sifat bilangan berpangkat:

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

Perhatikan perhitungan berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open parentheses 13 squared close parentheses to the power of begin display style 1 fourth end style end exponent cross times open parentheses 14 to the power of 5 close parentheses to the power of begin display style 2 over 15 end style end exponent over denominator open parentheses 13 to the power of begin display style 3 over 2 end style end exponent close parentheses to the power of begin display style 1 third end style end exponent cross times open parentheses 14 to the power of begin display style 1 fifth end style end exponent close parentheses to the power of begin display style 4 over 3 end style end exponent end fraction end cell equals cell fraction numerator 13 to the power of begin display style 2 cross times 1 fourth end style end exponent cross times 14 to the power of begin display style 5 cross times 2 over 15 end style end exponent over denominator 13 to the power of begin display style 3 over 2 cross times 1 third end style end exponent cross times 14 to the power of begin display style 1 fifth cross times 4 over 3 end style end exponent end fraction end cell row blank equals cell fraction numerator 13 to the power of begin display style 1 half end style end exponent cross times 14 to the power of begin display style 2 over 3 end style end exponent over denominator 13 to the power of begin display style 1 half end style end exponent cross times 14 to the power of begin display style 4 over 15 end style end exponent end fraction end cell row blank equals cell 13 to the power of 1 half minus 1 half end exponent cross times 14 to the power of 2 over 3 minus 4 over 15 end exponent end cell row blank equals cell 13 to the power of 0 cross times 14 to the power of fraction numerator 10 minus 4 over denominator 15 end fraction end exponent end cell row blank equals cell 1 cross times 14 to the power of 6 over 15 end exponent end cell row blank equals cell 14 to the power of 2 over 5 end exponent end cell end table

Dengan demikian hasil dari table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open parentheses 13 squared close parentheses to the power of begin display style 1 fourth end style end exponent cross times open parentheses 14 to the power of 5 close parentheses to the power of begin display style 2 over 15 end style end exponent over denominator open parentheses 13 to the power of begin display style 3 over 2 end style end exponent close parentheses to the power of begin display style 1 third end style end exponent cross times open parentheses 14 to the power of begin display style 1 fifth end style end exponent close parentheses to the power of begin display style 4 over 3 end style end exponent end fraction end cell equals cell 14 to the power of 2 over 5 end exponent end cell end table .

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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