Roboguru

Nilai dari 4−6(3×2−11+5×2−11)​=....

Pertanyaan

Nilai dari begin mathsize 14px style fraction numerator open parentheses 3 cross times 2 to the power of negative 11 end exponent plus 5 cross times 2 to the power of negative 11 end exponent close parentheses over denominator 4 to the power of negative 6 end exponent end fraction equals.... end style  

  1. 8undefined 

  2. 10undefined 

  3. 12undefined 

  4. 16undefined 

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator begin mathsize 14px style left parenthesis 3 cross times 2 to the power of negative 11 end exponent plus 5 cross times 2 to the power of negative 11 end exponent right parenthesis end style over denominator size 14px 4 to the power of size 14px minus size 14px 6 end exponent end fraction end cell size 14px equals cell fraction numerator size 14px 8 size 14px cross times size 14px 2 to the power of size 14px minus size 14px 11 end exponent over denominator begin mathsize 14px style left parenthesis 2 squared right parenthesis end style to the power of size 14px minus size 14px 6 end exponent end fraction end cell row blank size 14px equals cell fraction numerator size 14px 8 size 14px cross times size 14px 2 to the power of size 14px minus size 14px 11 end exponent over denominator begin mathsize 14px style 2 to the power of negative 12 end exponent end style end fraction end cell row blank size 14px equals cell size 14px 8 size 14px cross times size 14px 2 to the power of size 14px minus size 14px 11 size 14px plus size 14px 12 end exponent end cell row blank size 14px equals cell size 14px 8 size 14px cross times size 14px 2 end cell row blank size 14px equals size 14px 16 end table

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Lestari

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

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 (p9q−1p5q−5​)−2 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.

0

Roboguru

Nilai dari 27−34​+81−43​9−23​×3+27−1​adalah ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 9 to the power of negative begin display style 3 over 2 end style end exponent cross times 3 plus 27 to the power of negative 1 end exponent over denominator 27 to the power of negative begin display style 4 over 3 end style end exponent plus 81 to the power of negative begin display style 3 over 4 end style end exponent end fraction end cell equals cell fraction numerator open parentheses 3 squared close parentheses to the power of negative begin display style 3 over 2 end style plus 1 end exponent plus open parentheses 3 cubed close parentheses to the power of negative 1 end exponent over denominator open parentheses 3 cubed close parentheses to the power of negative begin display style 4 over 3 end style end exponent plus open parentheses 3 to the power of 4 close parentheses to the power of negative begin display style 3 over 4 end style end exponent end fraction end cell row blank equals cell fraction numerator 3 to the power of negative 3 plus 1 end exponent plus 3 to the power of negative 3 end exponent over denominator 3 to the power of negative 4 end exponent plus 3 to the power of negative 3 end exponent end fraction end cell row blank equals cell fraction numerator 3 to the power of negative 2 end exponent plus 3 to the power of negative 3 end exponent over denominator 3 to the power of negative 4 end exponent plus 3 to the power of negative 3 end exponent end fraction end cell row blank equals cell fraction numerator begin display style 1 over 9 end style plus begin display style 1 over 27 end style over denominator begin display style 1 over 81 end style plus begin display style 1 over 27 end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 3 plus 1 over denominator 27 end fraction end style over denominator begin display style fraction numerator 1 plus 3 over denominator 81 end fraction end style end fraction end cell row blank equals cell fraction numerator up diagonal strike 4 over denominator up diagonal strike 27 end fraction times fraction numerator begin display style stack up diagonal strike 81 with 3 on top end style over denominator up diagonal strike 4 end fraction end cell row blank equals 3 row blank blank blank end table end style 

Jadi, jawaban yang tepat adalah E

0

Roboguru

Bentuk sederhana dari (p9q−1p5q−5​)−2 adalah ...

Pembahasan Soal:

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 open parentheses fraction numerator p to the power of 5 times left parenthesis negative 2 right parenthesis end exponent times q to the power of left parenthesis negative 5 right parenthesis times left parenthesis negative 2 right parenthesis end exponent over denominator p to the power of 9 times left parenthesis negative 2 right parenthesis end exponent times q to the power of left parenthesis negative 1 right parenthesis times left parenthesis negative 2 right parenthesis end exponent end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator p to the power of negative 10 end exponent times q to the power of 10 over denominator p to the power of negative 18 end exponent times q squared end fraction close parentheses end cell row blank equals cell open parentheses p to the power of left parenthesis negative 10 right parenthesis plus 18 end exponent times q to the power of 10 minus 2 end exponent close parentheses end cell row blank equals cell open parentheses p to the power of 8 times q to the power of 8 close parentheses end cell row blank equals cell open parentheses p times q close parentheses to the power of 8 end cell end table


Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Sebutkan sifat-sifat eksponen.

Pembahasan Soal:

Sifat-sifat eksponen atau bilangan berpangkat diantaranya sebagai berikut:

  • 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 
  • 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 
  • 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 
  • 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 
  • open parentheses a over b close parentheses to the power of m equals a to the power of m over b to the power of m 
  • 1 over a to the power of m equals a to the power of negative m end exponent 
  • n-th root of a to the power of m end root equals a to the power of m over n end exponent 
  • a to the power of 0 equals 1 
  • 1 to the power of m equals 1 

Dengan demikian, diperoleh sifat-sifat eksponen seperti di atas.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved