Pertanyaan

Hasil dari adalah ....

Hasil dari $begin mathsize 14px style 3 open parentheses p to the power of 28 q to the power of negative 12 end exponent close parentheses to the power of 1 fourth end exponent minus 6 cube root of p to the power of 21 q to the power of negative 9 end exponent end root plus fraction numerator square root of cube root of p to the power of 42 end root end root over denominator q cubed end fraction end style$ adalah ....

$begin mathsize 14px style negative fraction numerator 4 p to the power of 7 over denominator q cubed end fraction end style$
$begin mathsize 14px style negative fraction numerator 2 p to the power of 7 over denominator q cubed end fraction end style$

H. Firmansyah

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Perhatikan bahwa Jadi, hasil dari adalah . Dengan demikian, jawaban yang tepat adalah B.

Perhatikan bahwa

$begin mathsize 12px style 3 open parentheses p to the power of 28 q to the power of negative 12 end exponent close parentheses to the power of 1 fourth end exponent minus 6 cube root of p to the power of 21 q to the power of negative 9 end exponent end root plus fraction numerator square root of cube root of p to the power of 42 end root end root over denominator q cubed end fraction equals 3 open parentheses p to the power of 28 q to the power of negative 12 end exponent close parentheses to the power of 1 fourth end exponent minus 6 open parentheses p to the power of 21 q to the power of negative 9 end exponent close parentheses to the power of 1 third end exponent plus fraction numerator square root of open parentheses p to the power of 42 close parentheses to the power of 1 third end exponent end root over denominator q cubed end fraction blank equals 3 open parentheses p to the power of 28 cross times 1 fourth end exponent q to the power of negative 12 cross times 1 fourth end exponent close parentheses minus 6 open parentheses p to the power of 21 cross times 1 third end exponent q to the power of negative 9 cross times 1 third end exponent close parentheses plus q to the power of negative 3 end exponent square root of open parentheses p to the power of 42 cross times 1 third end exponent close parentheses end root equals 3 p to the power of 7 q to the power of negative 3 end exponent minus 6 p to the power of 7 q to the power of negative 3 end exponent plus q to the power of negative 3 end exponent square root of open parentheses p to the power of 14 close parentheses end root equals 3 p to the power of 7 q to the power of negative 3 end exponent minus 6 p to the power of 7 q to the power of negative 3 end exponent plus q to the power of negative 3 end exponent open parentheses open parentheses p to the power of 14 close parentheses to the power of 1 half end exponent close parentheses equals 3 p to the power of 7 q to the power of negative 3 end exponent minus 6 p to the power of 7 q to the power of negative 3 end exponent plus q to the power of negative 3 end exponent p to the power of 7 equals open parentheses 3 minus 6 plus 1 close parentheses p to the power of 7 q to the power of negative 3 end exponent equals negative 2 p to the power of 7 q to the power of negative 3 end exponent equals negative fraction numerator 2 p to the power of 7 over denominator q cubed end fraction end style$

Jadi, hasil dari adalah .

Dengan demikian, jawaban yang tepat adalah B.