Roboguru

Nyatakan dalam bentuk pangkat negatif. g.

Pertanyaan

Nyatakan dalam bentuk pangkat negatif.

g. begin mathsize 14px style open parentheses 1 half close parentheses cubed end style 

Pembahasan Soal:

Ingat!

Sifat bilangan berpangkat

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over a to the power of m end cell equals cell a to the power of negative m end exponent end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of m close parentheses to the power of n end cell equals cell a to the power of m n end exponent end cell end table

Sehingga,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 half close parentheses cubed end cell equals cell open parentheses 2 to the power of negative 1 end exponent close parentheses cubed end cell row blank equals cell 2 to the power of open parentheses negative 1 cross times 3 close parentheses end exponent end cell row blank equals cell 2 to the power of negative 3 end exponent end cell end table end style

Dengan demikian, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 1 half close parentheses cubed end cell end table dinyatakan dalam pangkat negatif menjadi table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 2 to the power of negative 3 end exponent end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 13 Agustus 2021

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Pertanyaan yang serupa

Nilai dari  untuk ,  dan  adalah ....

Pembahasan Soal:

Nilai dari  

Diketahui begin mathsize 14px style p equals 27 end stylebegin mathsize 14px style q equals 4 end style dan begin mathsize 14px style r equals 3 end style, diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 p to the power of 2 over 3 end exponent q to the power of 1 half end exponent r to the power of negative 2 end exponent end cell equals cell 3 left parenthesis 27 right parenthesis to the power of 2 over 3 end exponent open parentheses 4 close parentheses to the power of fraction numerator 1 over denominator 2 end fraction end exponent open parentheses 3 close parentheses to the power of negative 2 end exponent end cell row blank equals cell 3 open parentheses 3 cubed close parentheses to the power of 2 over 3 end exponent open parentheses 2 squared close parentheses to the power of 1 half end exponent open parentheses 3 close parentheses to the power of negative 2 end exponent end cell row blank equals cell 3 open parentheses 3 squared close parentheses open parentheses 3 close parentheses to the power of negative 2 end exponent open parentheses 2 to the power of 1 close parentheses end cell row blank equals cell 3 open parentheses 3 to the power of 2 minus 2 end exponent close parentheses left parenthesis 2 to the power of 1 right parenthesis end cell row blank equals cell 3 times 1 times 2 end cell row blank equals 6 end table end style 

Jadi, jawaban yang tepat adalah B.

1

Roboguru

Hasil pemangkatan dari

Pembahasan Soal:

Berdasarkan sifat bilangan berpangkat yaitu a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent maka open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell open parentheses a to the power of 9 over 4 minus 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell row blank equals cell open parentheses a to the power of 6 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell end table

Berdasarkan sifat bilangan berpangkat open parentheses a to the power of m close parentheses to the power of n equals a to the power of m cross times n end exponent maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell open parentheses a to the power of 6 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell row blank equals cell a to the power of 6 over 4 cross times negative 5 end exponent end cell row blank equals cell a to the power of negative 30 over 4 end exponent end cell end table

Berdasarkan sifat bilangan berpangkat a to the power of negative n end exponent equals 1 over a to the power of n comma space a not equal to 0 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell a to the power of negative 30 over 4 end exponent end cell row blank equals cell 1 over a to the power of begin display style 30 over 4 end style end exponent end cell row blank equals cell 1 over a to the power of begin display style fraction numerator 28 plus 2 over denominator 4 end fraction end style end exponent end cell row blank equals cell 1 over a to the power of begin display style 28 over 4 end style plus begin display style 2 over 4 end style end exponent end cell row blank equals cell 1 over a to the power of 7 plus begin display style 2 over 4 end style end exponent end cell end table

Berdasarkan sifat bilangan berpangkat a to the power of m cross times a to the power of n equals a to the power of m plus n end exponent maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell 1 over a to the power of 7 plus begin display style 2 over 4 end style end exponent end cell row blank equals cell fraction numerator 1 over denominator a to the power of 7 cross times a to the power of begin display style 2 over 4 end style end exponent end fraction end cell end table

Berdasarkan sifat bilangan berpangkat pecahan yaitu a to the power of m over n end exponent equals n-th root of a to the power of m end root comma space a not equal to 0 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent end cell equals cell fraction numerator 1 over denominator a to the power of 7 cross times a to the power of begin display style 2 over 4 end style end exponent end fraction end cell row blank equals cell fraction numerator 1 over denominator a to the power of 7 cross times fourth root of a squared end root end fraction end cell end table

Jadi hasil pemangkatan dari open parentheses a to the power of 9 over 4 end exponent divided by a to the power of 3 over 4 end exponent close parentheses to the power of negative 5 end exponent adalah fraction numerator 1 over denominator a to the power of 7 space fourth root of a squared end root end fraction.

Oleh karena itu, jawaban yang benar adalah A.

9

Roboguru

Sederhanakan dan tulislah tanpa pangkat negatif. a.

Pembahasan Soal:

  • Menyederhanakan bilangan berpangkat dengan sifat berikut:

open parentheses a cross times b close parentheses to the power of m equals a to the power of m cross times b to the power of m a to the power of negative m end exponent equals 1 over a to the power of m 

Pembahasan:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 a close parentheses to the power of negative 2 end exponent end cell equals cell open parentheses 3 close parentheses to the power of negative 2 end exponent open parentheses a close parentheses to the power of negative 2 end exponent end cell row blank equals cell 1 over 3 squared 1 over a squared end cell row blank equals cell fraction numerator 1 over denominator 9 a squared end fraction end cell end table 

Dengan demikian, open parentheses 3 a close parentheses to the power of negative 2 end exponent equals fraction numerator 1 over denominator 9 a squared end fraction.

0

Roboguru

Sederhanakan bentuk-bentuk berikut! 3x6y2z12x2y−1z​

Pembahasan Soal:

Ingat kembali sifat bilangan berpangkat berikut.

am÷anam==amnam1,a=0

Sehingga, diperoleh perhitungan:

3x6y2z12x2y1z===4x26y12z114x4y3x4y34

Jadi, bentuk sederhana dari 3x6y2z12x2y1z adalahx4y34.

0

Roboguru

Tentukan hasil pemangkatan berikut!

Pembahasan Soal:

Ingat kembali aturan eksponen berikut ini!

  • a to the power of negative m end exponent equals 1 over a to the power of m 
  • a to the power of 1 equals a 
  • a to the power of 0 equals 1 

Dengan aturan eksponen di atas, diperoleh perhitungan sebagai berikut.

1.     Error converting from MathML to accessible text.
    table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell 3 to the power of 4 plus 3 cubed plus 3 squared plus 3 to the power of 1 plus 1 plus 1 over 3 to the power of 1 plus 1 over 3 squared end cell row blank equals cell 81 plus 27 plus 9 plus 3 plus 1 plus 1 third plus 1 over 9 end cell row blank equals cell 121 plus 1 third plus 1 over 9 end cell row blank equals cell 121 plus fraction numerator 3 plus 1 over denominator 9 end fraction end cell row blank equals cell 121 plus 4 over 9 end cell row blank equals cell 121 4 over 9 end cell end table 

2.     Error converting from MathML to accessible text. 
    table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell 1 plus 8 plus 27 plus 64 plus 124 plus 216 plus 343 plus 512 end cell row blank equals cell 1.295 end cell end table  

Jadi, hasil pemangkatan bilangan-bilangan tersebut adalah
(1) table attributes columnalign right center left columnspacing 0px end attributes row blank blank 121 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 4 over 9 end cell end table; dan (2) table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 295 end table 

1

Roboguru

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