Bentuk sederhana dari (24a5b−3c3a−2bc−3)−1 adalah ....

Pertanyaan

Bentuk sederhana dari $open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent$ adalah ....

$fraction numerator 8 straight a to the power of 7 straight c to the power of 4 over denominator straight b to the power of 4 end fraction$
$fraction numerator 8 straight a to the power of 10 straight c cubed over denominator straight b to the power of 4 end fraction$
$fraction numerator 8 straight a to the power of 7 straight c cubed over denominator straight b cubed end fraction$
$fraction numerator 8 straight a to the power of 10 straight b cubed over denominator straight c cubed end fraction$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

Diketahui suatu bentuk eksponen $open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent$ . Berdasarkan sifat bilangan berpangkat negatif yaitu a to the power of negative n end exponent equals 1 over a to the power of n maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses end fraction end cell row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator 1 over denominator fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction end fraction end cell row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c over denominator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent end fraction end cell end table$

Berdasarkan sifat bilangan berpangkat pembagian yaitu a to the power of m over a to the power of n equals a to the power of m minus n end exponent sehingga diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator begin display style stack up diagonal strike 24 with 8 on top end style straight a to the power of 5 straight b to the power of negative 3 end exponent straight c over denominator up diagonal strike 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent end fraction end cell row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell 8 a to the power of 5 minus open parentheses negative 2 close parentheses end exponent b to the power of negative 3 minus 1 end exponent c to the power of 1 minus open parentheses negative 3 close parentheses end exponent end cell row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell 8 a to the power of 7 b to the power of negative 4 end exponent c to the power of 4 end cell row cell open parentheses fraction numerator 3 straight a to the power of negative 2 end exponent bc to the power of negative 3 end exponent over denominator 24 straight a to the power of 5 straight b to the power of negative 3 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator 8 straight a to the power of 7 straight c to the power of 4 over denominator b to the power of 4 end fraction end cell row blank blank blank row blank blank blank end table$

Oleh karena itu, jawaban yang benar adalah A.

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