Roboguru

Jika a=2 dan b=3, tentukan nilai dari : c. ​​1−1+ab−11​1​​

Pertanyaan

Jika a=2 dan b=3, tentukan nilai dari :

c. 11+ab111

 

Pembahasan Soal:

Ingat!

Sifat bilangan berangkat:

  • an=an1     

Sehingga:

11+ab111==========11+2×311111+2×311111+3211133+211135111531553152125221 

Dengan demikian, nilai dari 11+ab111 adalah 221.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 30 Agustus 2021

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Pertanyaan yang serupa

Diketahui , dan . Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Ingat kembali:

straight a to the power of straight m over straight a to the power of straight n equals straight a to the power of straight m minus straight n end exponent straight a to the power of negative straight n end exponent equals 1 over straight a to the power of straight n straight a to the power of 0 equals 1 

Diketahui:

open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent 

Maka:

open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent equals open parentheses 12 over 84 times straight a to the power of negative 5 minus 2 end exponent times straight b to the power of 1 minus open parentheses negative 1 close parentheses end exponent times straight c to the power of 1 minus 1 end exponent close parentheses to the power of negative 1 end exponent open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent equals open parentheses 1 over 7 times straight a to the power of negative 7 end exponent times straight b squared times straight c to the power of 0 close parentheses to the power of negative 1 end exponent open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent equals open parentheses 1 over 7 close parentheses to the power of negative 1 end exponent times open parentheses straight a to the power of negative 7 end exponent close parentheses to the power of negative 1 end exponent times open parentheses straight b squared close parentheses to the power of negative 1 end exponent times open parentheses straight c to the power of 0 close parentheses to the power of negative 1 end exponent open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent equals 7 times straight a to the power of 7 times straight b to the power of negative 2 end exponent times straight c to the power of 0 open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent equals 7 times straight a to the power of 7 times 1 over straight b squared times 1 open parentheses fraction numerator 12 straight a to the power of negative 5 end exponent bc over denominator 84 straight a squared straight b to the power of negative 1 end exponent straight c end fraction close parentheses to the power of negative 1 end exponent equals fraction numerator 7 straight a to the power of 7 over denominator straight b squared end fraction 

Jadi, jawaban yang benar adalah D.

Roboguru

Jika  dan , nilai  adalah ....

Pembahasan Soal:

Sifat bilangan berpangkatan yang digunakan:
open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent 

Dengan menerapkan sifat pemangkatan bilangan berpangkat, diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 5 to the power of 10 x end exponent end cell equals 1600 row cell 5 to the power of 2 times 5 x end exponent end cell equals 1600 row cell open parentheses 5 to the power of 5 x end exponent close parentheses squared end cell equals 1600 row cell square root of open parentheses 5 to the power of 5 x end exponent close parentheses squared end root end cell equals cell square root of 1600 end cell row cell 5 to the power of 5 x end exponent end cell equals 40 end table end style 

Selanjutnya, dengan menerapkan sifat-sifat bilangan berpangkat, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 5 to the power of x minus 1 end exponent close parentheses to the power of 5 over 8 to the power of open parentheses negative square root of y close parentheses end exponent end cell equals cell 5 to the power of 5 open parentheses x minus 1 close parentheses end exponent over open parentheses 2 cubed close parentheses to the power of negative square root of y end exponent end cell row blank equals cell 5 to the power of 5 x minus 5 end exponent over 2 to the power of negative 3 square root of y end exponent end cell row blank equals cell fraction numerator begin display style 5 to the power of 5 x end exponent over 5 to the power of 5 end style over denominator begin display style 1 over 2 to the power of 3 square root of y end exponent end style end fraction end cell row blank equals cell 5 to the power of 5 x end exponent over 5 to the power of 5 cross times 2 to the power of 3 square root of y end exponent end cell row blank equals cell 5 to the power of 5 x end exponent over 5 to the power of 5 cross times open parentheses 2 to the power of square root of y end exponent close parentheses cubed end cell row blank equals cell 40 over 5 to the power of 5 cross times 25 cubed end cell row blank equals cell 40 over 5 to the power of 5 cross times open parentheses 5 squared close parentheses cubed end cell row blank equals cell 40 over 5 to the power of 5 cross times 5 to the power of 2 times 3 end exponent end cell row blank equals cell 40 over 5 to the power of 5 cross times 5 to the power of 6 end cell row blank equals cell 40 cross times 5 to the power of 6 minus 5 end exponent end cell row blank equals cell 40 cross times 5 to the power of 1 end cell row blank equals cell 40 cross times 5 end cell row blank equals 200 end table end style 

Jadi, nilai begin mathsize 14px style open parentheses 5 to the power of x minus 1 end exponent close parentheses to the power of 5 over 8 to the power of open parentheses negative square root of y close parentheses end exponent end style adalah 200.

Roboguru

Diketahui  dan , bentuk sederhana  adalah ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses fraction numerator 27 straight x to the power of negative 5 end exponent straight y to the power of negative 3 end exponent over denominator 3 to the power of 5 straight x to the power of negative 7 end exponent straight y to the power of negative 5 end exponent end fraction close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses 3 cubed over 3 to the power of 5 times straight x to the power of negative 5 minus open parentheses negative 7 close parentheses end exponent times straight y to the power of negative 3 minus open parentheses negative 5 close parentheses end exponent close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 3 to the power of 3 minus 5 end exponent times straight x squared times straight y squared close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 3 to the power of negative 2 end exponent times straight x squared times straight y squared close parentheses to the power of negative 1 end exponent end cell row blank equals cell 3 squared times straight x to the power of negative 2 end exponent times straight y to the power of negative 2 end exponent end cell row blank equals cell fraction numerator 9 over denominator straight x squared times straight y squared end fraction end cell row blank equals cell 9 over open parentheses xy close parentheses squared end cell end table end style 
 

Jadi, jawaban yang benar adalah E

Roboguru

Tentukan hasil pemangkatan bilangan-bilangan berikut! 1.

Pembahasan Soal:

Sebelum menjawab soal, ingatlah sifat-sifat bilangan berpangkat berikut.

left parenthesis straight i right parenthesis space 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 left parenthesis ii right parenthesis space a to the power of negative m end exponent equals 1 over a to the power of m

Berdasarkan sifat-sifat di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 squared close parentheses to the power of negative 3 end exponent end cell equals cell 3 to the power of 2 cross times open parentheses negative 3 close parentheses end exponent end cell row blank equals cell 3 to the power of negative 6 end exponent end cell row blank equals cell 1 over 3 to the power of 6 end cell row blank equals cell 1 over 729 end cell end table

Dengan demikian, hasil dari open parentheses 3 squared close parentheses to the power of negative 3 end exponent adalah 1 over 729.

Roboguru

Nyatakan bilangan berikut dengan meggunakan pangkat positif!

Pembahasan Soal:

Ingat kembali sifat bilangan berpangkat untuk pangkat negatif berikut.

a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0

Berdasarkan sifat di atas, maka diperoleh perhitungan berikut ini.

negative 12 to the power of negative 10 end exponent equals fraction numerator 1 over denominator negative 12 to the power of 10 end fraction

Jadi, bilangan negative 12 to the power of negative 10 end exponent dinyatakan dengan pangkat positif adalah fraction numerator 1 over denominator negative 12 to the power of 10 end fraction.

Roboguru

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