Roboguru

Sederhanakan h. 4d3×2d2:d4=...

Pertanyaan

Sederhanakan

h. 4d3×2d2:d4=...

Pembahasan Soal:

Ingat!

Sifat bilangan berangkat:

  • am×an=am+n
  • am:an=amn  

Sehingga:

4d3×2d2:d4====d44d3×2d2d48d3+2d48d58d 

Dengan demikian, bentuk sederhana dari 4d3×2d2:d4 adalah 8d.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Jika  dan , nilai dari  adalah ....

Pembahasan Soal:

Ingat!

Sifat perpangkatan antara lain:

  1. 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 
  2. 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 
  3. a to the power of m space colon space a to the power of n equals a to the power of m minus n end exponent 
  4. a to the power of negative m end exponent equals 1 over a to the power of m space d e n g a n space a to the power of m not equal to 0 

Untuk begin mathsize 14px style a equals 4 end style dan begin mathsize 14px style b equals 5 end style, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 4 to the power of 5 open parentheses 4 to the power of negative 2 end exponent times 5 close parentheses over denominator open parentheses 4 times 5 close parentheses squared end fraction end cell equals cell fraction numerator 4 to the power of 5 times 4 to the power of negative 2 end exponent times 5 over denominator 4 squared times 5 squared end fraction space bold left parenthesis bold sifat bold space bold 1 bold right parenthesis end cell row blank equals cell fraction numerator 4 to the power of 5 plus left parenthesis negative 2 right parenthesis end exponent times 5 over denominator 4 squared times 5 squared end fraction space bold left parenthesis bold sifat bold space bold 2 bold right parenthesis end cell row blank equals cell fraction numerator 4 cubed times 5 over denominator 4 squared times 5 squared end fraction end cell row blank equals cell 4 to the power of 3 minus 2 end exponent times 5 to the power of 1 minus 2 end exponent space bold left parenthesis bold sifat bold space bold 3 bold right parenthesis end cell row blank equals cell 4 times 5 to the power of negative 1 end exponent end cell row blank equals cell 4 times 1 fifth space bold left parenthesis bold sifat bold space bold 4 bold right parenthesis end cell row blank equals cell 4 over 5 end cell end table 

Jadi, begin mathsize 14px style fraction numerator a to the power of 5 open parentheses a to the power of negative 2 end exponent b close parentheses over denominator open parentheses a b close parentheses squared end fraction end styleequals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 4 over 5 end cell end table.

0

Roboguru

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 27 to the power of begin display style 2 over 3 end style end exponent over 9 to the power of begin display style 3 over 2 end style end exponent end cell equals cell open parentheses 3 cubed close parentheses to the power of begin display style 2 over 3 end style end exponent over open parentheses 3 squared close parentheses to the power of begin display style 3 over 2 end style end exponent end cell row blank equals cell 3 squared over 3 cubed end cell row blank equals cell 3 to the power of 2 minus 3 end exponent end cell row blank equals cell 3 to the power of negative 1 end exponent end cell end table end style 

Jadi, jawaban yang benar adalah A.

1

Roboguru

Sederhanakanlah persamaan berikut !

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator straight a cubed times straight b to the power of negative 2 end exponent over denominator straight a squared times straight b to the power of negative 5 end exponent times straight c to the power of negative 3 end exponent end fraction end cell equals cell straight a to the power of 3 minus 2 end exponent times straight b to the power of negative 2 plus 5 end exponent times straight c cubed end cell row blank equals cell straight a times straight b cubed times straight c cubed end cell end table


Jadi bentuk sederhana dari fraction numerator straight a cubed times straight b to the power of negative 2 end exponent over denominator straight a squared times straight b to the power of negative 5 end exponent times straight c to the power of negative 3 end exponent end fraction adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight a times straight b cubed times straight c cubed end cell end table.

0

Roboguru

Tentukan bentuk sederhana dari:

Pembahasan Soal:

Ingat bahwa:

bullet space a to the power of b cross times a to the power of c equals a to the power of b plus c end exponent bullet space fraction numerator a over denominator b over c end fraction equals fraction numerator a cross times thin space c over denominator b end fraction bullet space x to the power of a over x to the power of b equals x to the power of a minus b end exponent 

Mencari fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction end cell equals cell fraction numerator 10 cubed 10 squared over denominator 10 squared over 10 end fraction end cell row blank blank cell fraction numerator 10 cubed 10 squared over denominator 10 squared over 10 end fraction end cell row blank equals cell fraction numerator 10 to the power of 5 10 over denominator 10 squared end fraction end cell row blank equals cell 10 times 10 to the power of 5 minus 2 end exponent end cell row blank equals cell 10 cubed 10 end cell row blank equals cell 10 to the power of 3 plus 1 end exponent end cell row blank equals cell 10 to the power of 4 end cell row blank equals cell 10.000 end cell end table 

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 10 cubed cross times 10 squared over denominator 10 squared space colon space 10 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 10 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 000 end table.

0

Roboguru

Tentukan hasil dari operasi bilangan berpangkat berikut.

Pembahasan Soal:

Dengan menggunakan sifat pembagian bilangan berpangkat, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 4 m to the power of 4 n cubed over denominator 3 n p end fraction colon fraction numerator 4 m squared n cubed over denominator 3 n squared x squared end fraction end cell equals cell fraction numerator 4 m to the power of 4 n cubed over denominator 3 n p end fraction cross times fraction numerator 3 n squared x squared over denominator 4 m squared n cubed end fraction end cell row blank equals cell fraction numerator m squared n x squared over denominator p end fraction end cell end table 

Dengan demikian, hasil dari operasi bilangan berpangkat pada soal tersebut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator m squared n x squared over denominator p end fraction end cell end table. 

1

Roboguru

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