Bentuk sederhana dari (x−3y−1z5x4yz−3)−2 adalah ...

Pertanyaan

Bentuk sederhana dari $open parentheses fraction numerator x to the power of 4 y z to the power of negative 3 end exponent over denominator x to the power of negative 3 end exponent y to the power of negative 1 end exponent z to the power of 5 end fraction close parentheses to the power of negative 2 end exponent$ adalah ...

$fraction numerator z to the power of 16 over denominator x to the power of 12 y to the power of 4 end fraction$
$fraction numerator x to the power of 12 y to the power of 4 over denominator z to the power of 16 end fraction$
$fraction numerator x to the power of 12 y to the power of 16 over denominator y to the power of 4 end fraction$
$fraction numerator x to the power of 12 over denominator y to the power of 4 z to the power of 16 end fraction$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

tidak ada pilihan jawaban yang benar. Jawaban yang benar adalah $fraction numerator z to the power of 16 over denominator x to the power of 14 y to the power of 4 end fraction$

Pembahasan

Ingat bahwa 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 a to the power of m right parenthesis to the power of n equals a to the power of m n end exponent , dan a to the power of negative n end exponent equals 1 over a to the power of n maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator x to the power of 4 y z to the power of negative 3 end exponent over denominator x to the power of negative 3 end exponent y to the power of negative 1 end exponent z to the power of 5 end fraction close parentheses to the power of negative 2 end exponent end cell equals cell left parenthesis x to the power of 4 minus left parenthesis negative 3 right parenthesis end exponent y to the power of 1 minus left parenthesis negative 1 right parenthesis end exponent z to the power of negative 3 minus 5 end exponent right parenthesis to the power of negative 2 end exponent space end cell row blank equals cell left parenthesis x to the power of 7 y squared z to the power of negative 8 end exponent right parenthesis to the power of negative 2 end exponent space end cell row blank equals cell x to the power of negative 14 end exponent y to the power of negative 4 end exponent z to the power of 16 end cell row blank equals cell fraction numerator z to the power of 16 over denominator x to the power of 14 y to the power of 4 end fraction end cell end table$

Oleh karena itu, tidak ada pilihan jawaban yang benar. Jawaban yang benar adalah $fraction numerator z to the power of 16 over denominator x to the power of 14 y to the power of 4 end fraction$