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Jika diketahui  dan , maka nilai  adalah ....

Pertanyaan

Jika diketahui x equals 1 dan y equals 2, maka nilai fraction numerator open parentheses x to the power of negative 2 end exponent minus y to the power of negative 1 end exponent close parentheses over denominator open parentheses x to the power of negative 1 end exponent minus y to the power of negative 2 end exponent close parentheses end fraction adalah ....space 

  1. 1 third 

  2. 1 half 

  3. 2 over 3 

  4. 3 over 2  

  5. 5 over 2 

Pembahasan Soal:

Ingat kembali sifat bilangan berpangkat:

table attributes columnalign right center left columnspacing 0px end attributes row cell a to the power of negative m end exponent end cell equals cell 1 over a to the power of m end cell end table dengan a not equal to 0

Dengan demikian, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator open parentheses x to the power of negative 2 end exponent minus y to the power of negative 1 end exponent close parentheses over denominator open parentheses x to the power of negative 1 end exponent minus y to the power of negative 2 end exponent close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 1 over x squared end style minus begin display style 1 over y to the power of 1 end style close parentheses over denominator open parentheses begin display style 1 over x to the power of 1 end style minus begin display style 1 over y squared end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 1 over x squared minus 1 over y end style close parentheses over denominator open parentheses begin display style 1 over x minus 1 over y squared end style close parentheses end fraction space space space midline horizontal ellipsis open parentheses 1 close parentheses end cell end table 

Lalu, dengan menyubtitusikan x equals 1 dan y equals 2 ke open parentheses 1 close parentheses, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator open parentheses begin display style 1 over 1 squared minus 1 half end style close parentheses over denominator open parentheses begin display style 1 over 1 minus 1 over 2 squared end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 1 over 1 minus 1 half end style close parentheses over denominator open parentheses begin display style 1 over 1 minus 1 fourth end style close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 1 minus 1 half end style close parentheses over denominator open parentheses begin display style 1 minus 1 fourth end style close parentheses end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style over denominator begin display style 3 over 4 end style end fraction end cell row blank equals cell 1 half cross times 4 over 3 end cell row blank equals cell 2 over 3 end cell end table

Jadi, nilai fraction numerator open parentheses x to the power of negative 2 end exponent minus y to the power of negative 1 end exponent close parentheses over denominator open parentheses x to the power of negative 1 end exponent minus y to the power of negative 2 end exponent close parentheses end fraction adalah 2 over 3.

Oleh karena itu, jawaban yang benar adalah C.space 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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