Pertanyaan

Hasil dari adalah ....

Hasil dari 24 cube root of open parentheses 64 x to the power of 15 y to the power of negative 6 end exponent z squared close parentheses to the power of 1 half end exponent end root divided by 8 square root of open parentheses x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 close parentheses to the power of 1 third end exponent end root adalah ....

$begin mathsize 14px style fraction numerator 6 y cubed z squared over denominator x to the power of 4 end fraction end style$
$begin mathsize 14px style fraction numerator 6 x cubed y to the power of 4 over denominator z squared end fraction end style$
$begin mathsize 14px style fraction numerator 6 y to the power of 4 over denominator x cubed z squared end fraction end style$

R. Tri

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Perhatikan perhitungan berikut! Dengan demikian, hasil dari adalah . Jadi, jawaban yang tepat adalah C.

Perhatikan perhitungan berikut!

$begin mathsize 12px style 24 cube root of open parentheses 64 x to the power of 15 y to the power of negative 6 end exponent z squared close parentheses to the power of 1 half end exponent end root divided by 8 square root of open parentheses x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 close parentheses to the power of 1 third end exponent end root equals fraction numerator 24 cube root of open parentheses 64 x to the power of 15 y to the power of negative 6 end exponent z squared close parentheses to the power of 1 half end exponent end root over denominator 8 square root of open parentheses x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 close parentheses to the power of begin display style 1 third end style end exponent end root end fraction equals fraction numerator 3 open parentheses open parentheses 64 x to the power of 15 y to the power of negative 6 end exponent z squared close parentheses to the power of 1 half end exponent close parentheses to the power of begin display style 1 third end style end exponent over denominator open parentheses open parentheses x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 close parentheses to the power of begin display style 1 third end style end exponent close parentheses to the power of begin display style 1 half end style end exponent end fraction equals fraction numerator 3 open parentheses 64 x to the power of 15 y to the power of negative 6 end exponent z squared close parentheses to the power of 1 over 6 end exponent over denominator open parentheses x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 close parentheses to the power of begin display style 1 over 6 end style end exponent end fraction equals fraction numerator 3 root index 6 of 64 x to the power of 15 y to the power of negative 6 end exponent z squared end root over denominator root index 6 of x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 end root end fraction equals 3 root index 6 of fraction numerator 64 x to the power of 15 y to the power of negative 6 end exponent z squared over denominator x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 end fraction end root equals 3 root index 6 of 64 x to the power of 15 minus left parenthesis negative 3 right parenthesis end exponent y to the power of negative 6 minus 18 end exponent z to the power of 2 minus 14 end exponent end root equals 3 root index 6 of 2 to the power of 6 x to the power of 18 y to the power of negative 24 end exponent z to the power of negative 12 end exponent end root equals 3 times 2 times x cubed times y to the power of negative 4 end exponent times z to the power of negative 2 end exponent equals fraction numerator 6 x cubed over denominator y to the power of 4 z squared end fraction end style$

Dengan demikian, hasil dari 24 cube root of open parentheses 64 x to the power of 15 y to the power of negative 6 end exponent z squared close parentheses to the power of 1 half end exponent end root divided by 8 square root of open parentheses x to the power of negative 3 end exponent y to the power of 18 z to the power of 14 close parentheses to the power of 1 third end exponent end root adalah $begin mathsize 14px style fraction numerator 6 x cubed over denominator y to the power of 4 z squared end fraction end style$ .

Jadi, jawaban yang tepat adalah C.