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Pertanyaan

Bentuk sederhana dari 24 p 9 q 2 r 7 8 p 2 q 6 r 5 ​ = ....

  1. fraction numerator q cubed over denominator 3 p to the power of 8 q squared r to the power of 7 end fraction

  2. fraction numerator q to the power of 4 over denominator 3 p to the power of 7 r squared end fraction

  3. fraction numerator q cubed over denominator 3 p to the power of 7 r squared end fraction

  4. fraction numerator q to the power of 4 over denominator 3 p to the power of 7 r end fraction

  5. fraction numerator 3 q to the power of 4 over denominator p to the power of 7 r squared end fraction

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Y. Laksmi

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

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Pembahasan

Gunakan space sifat minus sifat space perpangkatan space yaitu space colon  straight a to the power of straight p over straight a to the power of straight q equals straight a to the power of straight p minus straight q end exponent  straight a to the power of negative straight p end exponent equals 1 over straight a to the power of straight p    Maka space colon thin space  equals fraction numerator 8 straight p squared straight q to the power of 6 straight r to the power of 5 over denominator 24 straight p to the power of 9 straight q squared straight r to the power of 7 end fraction  equals 1 third straight p to the power of 2 minus 9 end exponent straight q to the power of 6 minus 2 end exponent straight r to the power of 5 minus 7 end exponent  equals 1 third straight p to the power of negative 7 end exponent straight q to the power of 4 straight r to the power of negative 2 end exponent  equals fraction numerator straight q to the power of 4 over denominator 3 straight p to the power of 7 straight r squared end fraction

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Bentuk sederhana dari ( 32 p 2 q − 1 r 6 4 p − 1 q 2 r 3 ​ ) − 1 adalah....

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