Pertanyaan

Hasil ( 6 x − 1 y 3 2 x 2 y ) 2 ÷ ( 3 x 4 y − 1 x y 3 ) 3 adalah ...

Hasil $(\frac{2 x ^{2} y}{6 x ^{- 1} y ^{3}})^{2} \div (\frac{x y ^{3}}{3 x ^{4} y ^{- 1}})^{3}$ adalah ...

$fraction numerator 3 x to the power of 15 over denominator y to the power of 16 end fraction$
$fraction numerator 2 x to the power of 15 over denominator 3 y to the power of 16 end fraction$
$fraction numerator x to the power of 15 over denominator 9 y to the power of 16 end fraction$
$fraction numerator 9 x to the power of 16 over denominator y to the power of 15 end fraction$
$fraction numerator x to the power of 16 over denominator 3 y to the power of 15 end fraction$

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Ingat kembali beberapasifat bilangan berpangkat berikut: Sehingga soal tersebut dapat kita hitung sebagai berikut: Jadi, . Oleh karena itu, jawaban yang benar adalah A.

Ingat kembali beberapa sifat bilangan berpangkat berikut:

Sehingga soal tersebut dapat kita hitung sebagai berikut:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 2 straight x squared straight y over denominator 6 straight x to the power of negative 1 end exponent straight y cubed end fraction close parentheses squared divided by open parentheses fraction numerator xy cubed over denominator 3 straight x to the power of 4 straight y to the power of negative 1 end exponent end fraction close parentheses cubed end cell equals cell open parentheses 1 third x to the power of open parentheses 2 plus 1 close parentheses end exponent cross times y to the power of open parentheses 1 minus 3 close parentheses end exponent close parentheses squared divided by open parentheses 1 third x to the power of open parentheses 1 minus 4 close parentheses end exponent cross times y to the power of open parentheses 3 plus 1 close parentheses end exponent close parentheses cubed end cell row blank equals cell open parentheses 1 third x cubed cross times y to the power of negative 2 end exponent close parentheses squared divided by open parentheses 1 third x to the power of negative 3 end exponent cross times y to the power of 4 close parentheses cubed end cell row blank equals cell open parentheses 1 over 9 x to the power of 3 cross times 2 end exponent cross times y to the power of negative 2 cross times 2 end exponent close parentheses divided by open parentheses 1 over 27 x to the power of negative 3 cross times 3 end exponent cross times y to the power of 4 cross times 3 end exponent close parentheses end cell row blank equals cell open parentheses 1 over 9 x to the power of 6 cross times y to the power of negative 4 end exponent close parentheses divided by open parentheses 1 over 27 x to the power of negative 9 end exponent cross times y to the power of 12 close parentheses end cell row blank equals cell fraction numerator open parentheses begin display style 1 over 9 end style x to the power of 6 cross times y to the power of negative 4 end exponent close parentheses over denominator open parentheses begin display style 1 over 27 end style x to the power of negative 9 end exponent cross times y to the power of 12 close parentheses end fraction end cell row blank equals cell 27 over 9 x to the power of 6 plus 9 end exponent cross times y to the power of negative 4 minus 12 end exponent end cell row blank equals cell 3 x to the power of 15 y to the power of negative 16 end exponent end cell row blank equals cell fraction numerator 3 x to the power of 15 over denominator y to the power of 16 end fraction end cell end table$

Jadi, $table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 2 straight x squared straight y over denominator 6 straight x to the power of negative 1 end exponent straight y cubed end fraction close parentheses squared divided by open parentheses fraction numerator xy cubed over denominator 3 straight x to the power of 4 straight y to the power of negative 1 end exponent end fraction close parentheses cubed end cell equals cell fraction numerator 3 x to the power of 15 over denominator y to the power of 16 end fraction end cell end table$ .

Oleh karena itu, jawaban yang benar adalah A.