Roboguru

Bentuk sederhana dari ​​(a−3.b4.c3)2(a4.b−3.c2)−1​​ adalah ...

Pertanyaan

Bentuk sederhana dari (a3.b4.c3)2(a4.b3.c2)1 adalah ...

  1. a.b.c5

  2. a2.b5.c8

  3. a7.b7.c1

  4. a2.b5.c8

  5. a10.b11.c4

Pembahasan Soal:

Ingat!

Sifat bilangan berangkat:

  • am:an=amn  
  • (a×b)n=an×bn 
  • (am)n=am×n   

Sehingga:

(a3.b4.c3)2(a4.b3.c2)1=====(a3)2.(b4)2.(c3)2(a4)1.(b3)1.(c2)1a3×2.b4×2.c3×2a4×(1).b3×(1).c2×(1)a6.b8.c6a4.b3.c2a4(6).b38.c26a2.b5.c8 

Oleh karena itu, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Bentuk sederhana dari  adalah ...

Pembahasan Soal:

Dengan menggunakan sifat-sifat dari bilangan berpangkat (eksponen), didapat:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of negative 3 end exponent b to the power of 5 c to the power of negative 8 end exponent close parentheses cubed over open parentheses a to the power of 4 b to the power of negative 3 end exponent c to the power of 10 close parentheses to the power of negative 2 end exponent end cell equals cell fraction numerator open parentheses a to the power of negative 9 end exponent b to the power of 15 c to the power of negative 24 end exponent close parentheses over denominator open parentheses a to the power of negative 8 end exponent b to the power of 6 c to the power of negative 20 end exponent close parentheses end fraction end cell row blank equals cell a to the power of negative 9 minus left parenthesis negative 8 right parenthesis end exponent b to the power of 15 minus 6 end exponent c to the power of negative 24 minus left parenthesis negative 20 right parenthesis end exponent end cell row blank equals cell a to the power of negative 1 end exponent b to the power of 9 c to the power of negative 4 end exponent end cell row blank equals cell fraction numerator b to the power of 9 over denominator a c to the power of 4 end fraction end cell end table end style 

Jadi, bentuk sederhananya adalah begin mathsize 14px style fraction numerator b to the power of 9 over denominator a c to the power of 4 end fraction end style.

Dengan demikian, jawaban yang tepat adalah C.

Roboguru

Sederhanakan dalam bentuk pangkat positif.

Pembahasan Soal:

Ingat kembali beberapa sifat bilangan berpangkat berikut:

  1. 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
  2. a to the power of negative m end exponent equals 1 over a to the power of m
  3. 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
  4. a to the power of m over a to the power of n equals a to the power of m minus n end exponent


Sehingga soal tersebut dapat kita hitung sebagai berikut:


open parentheses fraction numerator 4 x y squared over denominator 6 x cubed y end fraction close parentheses cubed cross times open parentheses fraction numerator 2 x cubed y over denominator 3 x squared y to the power of negative 2 end exponent end fraction close parentheses to the power of negative 2 end exponent equals open parentheses 2 over 3 x to the power of 1 minus 3 end exponent y to the power of 2 minus 1 end exponent close parentheses cubed cross times open parentheses 2 over 3 x to the power of 3 minus 2 end exponent y to the power of 1 plus 2 end exponent close parentheses to the power of negative 2 end exponent equals open parentheses 2 over 3 x to the power of negative 2 end exponent y close parentheses cubed cross times open parentheses 2 over 3 x y cubed close parentheses to the power of negative 2 end exponent equals open parentheses begin display style 2 over 3 x to the power of negative 2 end exponent y end style close parentheses cubed over open parentheses begin display style 2 over 3 x y cubed end style close parentheses squared equals fraction numerator open parentheses begin display style 2 over 3 end style close parentheses cubed x to the power of negative 6 end exponent y cubed over denominator open parentheses begin display style 2 over 3 end style close parentheses squared x squared y to the power of 6 end fraction equals open parentheses 2 over 3 close parentheses to the power of 3 minus 2 end exponent x to the power of negative 6 minus 2 end exponent y to the power of 3 minus 6 end exponent equals open parentheses 2 over 3 close parentheses x to the power of negative 8 end exponent y to the power of negative 3 end exponent equals fraction numerator 2 over denominator 3 x to the power of 8 y cubed end fraction


Jadi, open parentheses fraction numerator 4 x y squared over denominator 6 x cubed y end fraction close parentheses cubed cross times open parentheses fraction numerator 2 x cubed y over denominator 3 x squared y to the power of negative 2 end exponent end fraction close parentheses to the power of negative 2 end exponent equals fraction numerator 2 over denominator 3 x to the power of 8 y cubed end fraction.

Roboguru

Dengan menggunakan tanda , atau  nyatakan perbandingan masing-masing bilangan berikut. g.

Pembahasan Soal:

Untuk mengetahui hubungan 2 comma 713 to the power of 3 comma 14 end exponent space... space 3 comma 14 to the power of 2 comma 713 end exponent, maka kita menggunakan pemisalan sebagai berikut:

a to the power of b... b to the power of a comma space S y a r a t space a less than b

Kita misalkan a equals 4 space d a n space b equals 5, maka diperoleh:

table attributes columnalign right center left columnspacing 2px end attributes row cell 4 to the power of 5 end cell cell... end cell cell 5 to the power of 4 end cell row cell 1.024 end cell cell... end cell 625 row cell 1.024 end cell greater than 625 end table

Sedemikian sehingga:

a to the power of b greater than b to the power of a comma space s y a r a t space a less than b

Jika a equals 2 comma 713 space & space b equals 3 comma 14 maka diperoleh:

2 comma 713 to the power of 3 comma 14 end exponent greater than 3 comma 14 to the power of 2 comma 713 end exponent dengan 2 comma 713 space less than space 3 comma 14.

Jadi, 2 comma 713 to the power of 3 comma 14 end exponent greater than 3 comma 14 to the power of 2 comma 713 end exponent.

Roboguru

Bentuk sederhana dari  adalah ... .

Pembahasan Soal:

Mencari bentuk sederhana dari open parentheses fraction numerator 12 a cubed b squared over denominator 4 a b to the power of 5 end fraction close parentheses to the power of negative 1 end exponent:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 12 a cubed b squared over denominator 4 a b to the power of 5 end fraction close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses 12 over 4 times fraction numerator a cubed b squared over denominator a b to the power of 5 end fraction close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 3 times a to the power of 3 minus 1 end exponent b to the power of 2 minus 5 end exponent close parentheses to the power of negative 1 end exponent end cell row blank equals cell open parentheses 3 a squared b to the power of negative 3 end exponent close parentheses to the power of negative 1 end exponent end cell row blank equals cell 3 to the power of negative 1 end exponent a to the power of negative 2 end exponent b cubed end cell row blank equals cell fraction numerator b cubed over denominator 3 a squared end fraction end cell end table 

Jadi, bentuk sederhana dari open parentheses fraction numerator 12 a cubed b squared over denominator 4 a b to the power of 5 end fraction close parentheses to the power of negative 1 end exponent adalah fraction numerator b cubed over denominator 3 a squared end fraction.

Dengan demikian, jawaban yang tepat adalah A.

Roboguru

Bentuk sederhana dari  adalah ...

Pembahasan Soal:

Mencari bentuk sederhana dari open parentheses fraction numerator p to the power of 5 q to the power of negative 5 end exponent over denominator p to the power of 9 q to the power of negative 1 end exponent end fraction close parentheses to the power of negative 2 end exponent:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator p to the power of 5 q to the power of negative 5 end exponent over denominator p to the power of 9 q to the power of negative 1 end exponent end fraction close parentheses to the power of negative 2 end exponent end cell equals cell 1 over open parentheses fraction numerator p to the power of 5 q to the power of negative 5 end exponent over denominator p to the power of 9 q to the power of negative 1 end exponent end fraction close parentheses squared end cell row blank equals cell fraction numerator 1 over denominator open parentheses fraction numerator p to the power of 10 q to the power of negative 10 end exponent over denominator p to the power of 18 q to the power of negative 2 end exponent end fraction close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator p to the power of negative 8 end exponent q to the power of negative 8 end exponent end fraction end cell row blank equals cell p to the power of 8 q to the power of 8 end cell row blank equals cell open parentheses p q close parentheses to the power of 8 end cell end table  

Jadi, bentuk sederhana dari open parentheses fraction numerator p to the power of 5 q to the power of negative 5 end exponent over denominator p to the power of 9 q to the power of negative 1 end exponent end fraction close parentheses to the power of negative 2 end exponent adalah open parentheses p q close parentheses to the power of 8.

Dengan demikian, jawaban yang tepat adalah A.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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