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Jika a=4 dan b=5, nilai dari (ab)2a5(a−2b)​ adalah ....

Pertanyaan

Jika begin mathsize 14px style a equals 4 end style dan begin mathsize 14px style b equals 5 end style, nilai dari 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 style 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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nyatakanlah bilangan berikut dalam pangkat positif. a. 43×4−5

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 cubed cross times 4 to the power of negative 5 end exponent end cell equals cell 4 to the power of 3 minus 5 end exponent end cell row blank equals cell 4 to the power of negative 2 end exponent end cell row blank equals cell 1 over 4 squared end cell row blank equals cell 1 over 2 to the power of 4 end cell row blank equals cell 1 over 16 end cell end table 

Jadi, bilangan tersebut dalam pangkat positif adalah 1 over 2 to the power of 4 space atau space 1 over 16.

0

Roboguru

Nilai 365×256−18−2×244​ adalah ...

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 cross times a to the power of n equals a to the power of m plus n end exponent
  5. 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:


table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 8 to the power of negative 2 end exponent cross times 24 to the power of 4 over denominator 36 to the power of 5 cross times 256 to the power of negative 1 end exponent end fraction end cell equals cell fraction numerator open parentheses 2 cubed close parentheses to the power of negative 2 end exponent cross times open parentheses 2 cubed cross times 3 close parentheses to the power of 4 over denominator open parentheses 2 squared cross times 3 squared close parentheses to the power of 5 cross times open parentheses 2 to the power of 8 close parentheses to the power of negative 1 end exponent end fraction end cell row blank equals cell fraction numerator 2 to the power of negative 6 end exponent cross times 2 to the power of 12 cross times 3 to the power of 4 over denominator 2 to the power of 10 cross times 3 to the power of 10 cross times 2 to the power of negative 8 end exponent end fraction end cell row blank equals cell fraction numerator 2 to the power of 12 minus 6 end exponent cross times 3 to the power of 4 over denominator 2 to the power of 10 minus 8 end exponent cross times 3 to the power of 10 end fraction end cell row blank equals cell fraction numerator 2 to the power of 6 cross times 3 to the power of 4 over denominator 2 squared cross times 3 to the power of 10 end fraction end cell row blank equals cell 2 to the power of 6 minus 2 end exponent cross times 3 to the power of 4 minus 10 end exponent end cell row blank equals cell 2 to the power of 4 cross times 3 to the power of negative 6 end exponent end cell row blank equals cell 2 to the power of 4 over 3 to the power of 6 end cell row blank equals cell 16 over 729 end cell end table


Jadi, fraction numerator 8 to the power of negative 2 end exponent cross times 24 to the power of 4 over denominator 36 to the power of 5 cross times 256 to the power of negative 1 end exponent end fraction equals 16 over 729.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Nyatakanlah bilangan berikut dalam pangkat positif. b. (41​)−3×44

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 fourth close parentheses to the power of negative 3 end exponent times 4 to the power of 4 end cell equals cell 1 over open parentheses 1 fourth close parentheses cubed cross times 4 to the power of 4 end cell row blank equals cell 4 cubed cross times 4 to the power of 4 end cell row blank equals cell 4 to the power of 3 plus 4 end exponent end cell row blank equals cell 4 to the power of 7 end cell row blank equals cell open parentheses 2 squared close parentheses to the power of 7 end cell row blank equals cell 2 to the power of 14 end cell row blank equals cell 16.384 end cell end table  

Jadi, bilangan tersebut dalam pangkat positif 2 to the power of 14 space atau space 16.384.

0

Roboguru

Dengan menggunakan tanda "" role="math" src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAEwAAAASCAYAAADxEzisAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAABGJhU0UAAAAKIPF7gAAAALVJREFUeNrtk7ENgzAQRW8thJglOyA2SssKE...

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.

1

Roboguru

Bentuk sederhana dari (4ab512a3b2​)−1 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.

0

Roboguru

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