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Hasil dari 29×4−3÷22 adalah ...

Pertanyaan

Hasil dari 2 to the power of 9 cross times 4 to the power of negative 3 end exponent divided by 2 squared adalah ...

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Pembahasan Video:

Pembahasan Soal:

Ingat operasi perkalian dan pembagian pada bilangan berpangkat dan juga sifat-sifat dari bilangan berpangkat sendiri.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of 9 cross times 4 to the power of negative 3 end exponent divided by 2 squared end cell equals cell 2 to the power of 9 cross times open parentheses 2 squared close parentheses to the power of negative 3 end exponent divided by 2 squared end cell row blank equals cell 2 to the power of 9 cross times 2 to the power of open parentheses 2 right parenthesis cross times left parenthesis negative 3 close parentheses end exponent divided by 2 squared end cell row blank equals cell 2 to the power of 9 cross times 2 to the power of negative 6 end exponent divided by 2 squared end cell row blank equals cell 2 to the power of 9 plus open parentheses negative 6 close parentheses minus 2 end exponent end cell row blank equals cell 2 to the power of 9 minus 6 minus 2 end exponent end cell row blank equals cell 2 to the power of 1 end cell row blank equals 2 end table

Oleh karena itu, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 08 Oktober 2021

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Pertanyaan yang serupa

Sederhanakan operasi bilangan berpangkat berikut! a. (a43​×a−83​÷a61​)8 b. (a21​×a−32​÷b61​)−6

Pembahasan Soal:

Ingatlah sifat-sifat bilangan berpangkat berikut.

left parenthesis straight i right parenthesis space 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 left parenthesis ii right parenthesis space 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 left parenthesis iii right parenthesis space open parentheses a divided by b close parentheses to the power of m equals a to the power of m divided by b to the power of m left parenthesis iv 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  

Berdasarkan sifat-sifat di atas, maka:

a)

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 3 over 4 end exponent cross times a to the power of negative 3 over 8 end exponent divided by a to the power of 1 over 6 end exponent close parentheses to the power of 8 end cell equals cell open parentheses a to the power of 3 over 4 plus open parentheses negative 3 over 8 close parentheses minus 1 over 6 end exponent close parentheses to the power of 8 end cell row blank equals cell open parentheses a to the power of fraction numerator 18 minus 9 minus 4 over denominator 24 end fraction end exponent close parentheses to the power of 8 end cell row blank equals cell open parentheses a to the power of 5 over 24 end exponent close parentheses to the power of 8 end cell row blank equals cell a to the power of 5 over 24 cross times 8 end exponent end cell row blank equals cell a to the power of 40 over 24 end exponent end cell row blank equals cell a to the power of 5 over 3 end exponent end cell end table

b)

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of 1 half end exponent cross times a to the power of negative 2 over 3 end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell equals cell open parentheses a to the power of 1 half plus open parentheses negative 2 over 3 close parentheses end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell open parentheses a to the power of fraction numerator 3 minus 4 over denominator 6 end fraction end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell open parentheses a to the power of negative 1 over 6 end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell open parentheses a to the power of negative 1 over 6 end exponent close parentheses to the power of negative 6 end exponent divided by open parentheses b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent end cell row blank equals cell a to the power of negative fraction numerator 1 over denominator up diagonal strike 6 end fraction cross times open parentheses negative up diagonal strike 6 close parentheses end exponent divided by b to the power of fraction numerator 1 over denominator up diagonal strike 6 end fraction cross times open parentheses negative up diagonal strike 6 close parentheses end exponent end cell row blank equals cell a divided by b to the power of negative 1 end exponent end cell row blank equals cell a divided by 1 over b end cell row blank equals cell a b end cell end table

Dengan demikian, bentuk sederhana dari  open parentheses a to the power of 3 over 4 end exponent cross times a to the power of negative 3 over 8 end exponent divided by a to the power of 1 over 6 end exponent close parentheses to the power of 8 equals a to the power of 5 over 3 end exponent, dan open parentheses a to the power of 1 half end exponent cross times a to the power of negative 2 over 3 end exponent divided by b to the power of 1 over 6 end exponent close parentheses to the power of negative 6 end exponent equals a b.

1

Roboguru

(b2)3a−2b5×ab−3​=...

Pembahasan Soal:

Ingat : 

  • am×an=am+n 
  • am÷an=amn 
  • (am)n=am×n 
  • am=am1 

Perhatikan perhitungan berikut 

(b2)3a2b5×ab3======b2×3a2+1×b5+(3)b6a1×b2a1×b26a1×b4a1×b41ab41 

Dengan demikian, (b2)3a2b5×ab3=ab41

0

Roboguru

Sederhanakan bentuk-bentuk berikut. b. 32(a4b2)2(9a3b)3×ab2​

Pembahasan Soal:

Ingat kembali sifat-sifat bilangan berpangkat berikut:

(am)n=am×naman=am+nanam=amn 

Bentuk sederhana dari 32(a4b2)2(9a3b)3×ab2 dapat ditentukan sebagai berikut:

32(a4b2)2(9a3b)3×ab2=======32(a4)2(b2)293(a3)3b3ab232a8b4(32)3a9b3+2a32a8b436a9+1b532a8b436a10b5362a108b5434a2b181a2b  

Jadi, bentuk sederhana dari 32(a4b2)2(9a3b)3×ab2 adalah 81a2b.

0

Roboguru

Jika (3p)2−(2p)2(3p2)2+(4p2)2​=80, nilai dari p2=....

Pembahasan Soal:

Dengan menerapkan sifat bilangan berpangkat, maka dapat dilakukan perhitungan seperti di bawah ini

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator begin mathsize 14px style left parenthesis 3 p squared right parenthesis squared plus left parenthesis 4 p squared right parenthesis squared end style over denominator begin mathsize 14px style left parenthesis 3 p right parenthesis squared minus left parenthesis 2 p right parenthesis squared end style end fraction end cell size 14px equals size 14px 80 row cell fraction numerator begin mathsize 14px style 9 p to the power of 4 plus 16 p to the power of 4 end style over denominator begin mathsize 14px style 9 p squared minus 4 p squared end style end fraction end cell size 14px equals size 14px 80 row cell fraction numerator begin mathsize 14px style 25 p to the power of 4 end style over denominator begin mathsize 14px style 5 p squared end style end fraction end cell size 14px equals size 14px 80 row cell size 14px 5 size 14px p to the power of size 14px 2 end cell size 14px equals size 14px 80 row cell size 14px p to the power of size 14px 2 end cell size 14px equals size 14px 16 end table

Jadi, jawaban yang tepat adalah B.

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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