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Hitung ( 5 3 × 8 a 2 ) 3 1 ​ × 2 − 5 , Beri jawaban dalam bentukindeks!

Hitung , Beri jawaban dalam bentuk indeks!

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N. Puspita

Master Teacher

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hasil dari adalah

hasil dari open parentheses 5 cubed cross times 8 a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent adalah fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 16 end fraction 

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Jadi, hasil dari adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 5 cubed cross times 8 a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent end cell equals cell open parentheses 5 cubed cross times 2 cubed a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent end cell row blank equals cell 5 cross times 2 a to the power of 2 over 3 end exponent cross times 2 to the power of negative 5 end exponent end cell row blank equals cell 5 cross times 2 to the power of negative 4 end exponent cross times a to the power of 2 over 3 end exponent end cell row blank equals cell fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 2 to the power of 4 end fraction end cell row blank equals cell fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 16 end fraction end cell end table

Jadi, hasil dari open parentheses 5 cubed cross times 8 a squared close parentheses to the power of 1 third end exponent cross times 2 to the power of negative 5 end exponent adalah fraction numerator 5 a to the power of begin display style 2 over 3 end style end exponent over denominator 16 end fraction 

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Tentukan nilai bentuk aljabar berikut jika nilai x = 16 dan y = 64 ! e. ( x y ) − 5 2 ​ f. 12 x − 2 1 ​ y 3 1 ​

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