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Pertanyaan

Jika bentuk 3 a b c 3 ​ ​ dapat disederhanakan menjadi a x b y c z , maka x − 1 + y − 1 − z − 1 = ....

Jika bentuk  dapat disederhanakan menjadi , maka  = ....

  1. -1

  2. 5

  3. 7

  4. 11

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

jawaban yang tepat adalah C.

Pembahasan

Dengan menggunakan sifat bilangan berpangkat, kita peroleh Pada soal diketahui bahwa dapat disederhanakan menjadi . Perhatikan bahwa sebelumnya kita telah memperoleh , sehingga didapat . Oleh karena itu, didapat Sehingga Jadi, jawaban yang tepat adalah C.

Dengan menggunakan sifat bilangan berpangkat,  kita peroleh

begin mathsize 14px style cube root of a square root of b c cubed end root end root equals open parentheses a square root of b c cubed end root close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space equals open parentheses a open parentheses b c cubed close parentheses to the power of 1 half end exponent close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space equals open parentheses a b to the power of 1 half end exponent open parentheses c cubed close parentheses to the power of 1 half end exponent close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space equals open parentheses a b to the power of 1 half end exponent c to the power of 3 open parentheses 1 half close parentheses end exponent close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space equals open parentheses a b to the power of 1 half end exponent c to the power of 3 over 2 end exponent close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space equals a to the power of 1 third end exponent open parentheses b to the power of 1 half end exponent close parentheses to the power of 1 third end exponent open parentheses c to the power of 3 over 2 end exponent close parentheses to the power of 1 third end exponent space space space space space space space space space space space space space space space equals a to the power of 1 third end exponent b to the power of 1 half open parentheses 1 third close parentheses end exponent c to the power of 3 over 2 open parentheses 1 third close parentheses end exponent space space space space space space space space space space space space space space space equals box enclose a to the power of 1 third end exponent b to the power of 1 over 6 end exponent c to the power of 1 half end exponent end enclose end style

Pada soal diketahui bahwa begin mathsize 14px style cube root of a square root of b c cubed end root end root end style dapat disederhanakan menjadi begin mathsize 14px style a to the power of x b to the power of y c to the power of z end style. Perhatikan bahwa sebelumnya kita telah memperoleh begin mathsize 14px style cube root of a square root of b c cubed end root end root equals a to the power of 1 third end exponent b to the power of 1 over 6 end exponent c to the power of 1 half end exponent end style, sehingga didapat

begin mathsize 14px style a to the power of x b to the power of y c to the power of z equals a to the power of 1 third end exponent b to the power of 1 over 6 end exponent c to the power of 1 half end exponent end style.

Oleh karena itu, didapat

begin mathsize 14px style x equals 1 third y equals 1 over 6 z equals 1 half end style   

Sehingga

begin mathsize 14px style x to the power of negative 1 end exponent plus y to the power of negative 1 end exponent minus z to the power of negative 1 end exponent equals 1 over x plus 1 over y minus 1 over z space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 1 over denominator 1 third end fraction plus fraction numerator 1 over denominator 1 over 6 end fraction minus fraction numerator 1 over denominator 1 half end fraction space space space space space space space space space space space space space space space space space space space space space space space space equals 3 plus 6 minus 2 space space space space space space space space space space space space space space space space space space space space space space space space equals box enclose 7 end style   

Jadi, jawaban yang tepat adalah C.

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