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Pertanyaan

open parentheses 2 plus square root of 3 close parentheses open parentheses square root of 4 minus square root of 3 close parentheses plus 1 third 

  1. 0 comma 3 

  2. 1 comma 3 

  3. 2 comma 3 

  4. 3 comma 3 

  5. 4 

Pembahasan Video:

Pembahasan Soal:

Ingat konsep perkalian bentuk akar:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of a minus square root of b close parentheses open parentheses square root of a plus square root of b close parentheses end cell equals cell square root of a squared end root minus square root of a b end root plus square root of a b end root minus square root of b squared end root end cell row blank equals cell a minus b end cell end table  

sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 plus square root of 3 close parentheses open parentheses square root of 4 minus square root of 3 close parentheses plus 1 third end cell equals cell open parentheses 2 plus square root of 3 close parentheses open parentheses 2 minus square root of 3 close parentheses plus 1 third end cell row blank equals cell 2 times 2 minus 2 square root of 3 plus 2 square root of 3 minus square root of 9 plus 1 third end cell row blank equals cell 4 minus 3 plus 1 third end cell row blank equals cell 1 plus 1 third end cell row blank equals cell 1 1 third end cell row blank equals cell 1 comma 3 end cell end table

Nilai dari open parentheses 2 plus square root of 3 close parentheses open parentheses square root of 4 minus square root of 3 close parentheses plus 1 third adalah 1 comma 3.

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 13 September 2021

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Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell x y end cell equals cell left parenthesis 1 plus square root of 3 right parenthesis left parenthesis 1 minus square root of 3 right parenthesis end cell row blank equals cell 1 minus square root of 3 plus square root of 3 minus 3 end cell row blank equals cell 1 minus 3 end cell row blank equals cell negative 2 end cell end table  

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Bentuk sederhana dari adalah

Pembahasan Soal:

Untuk menemukan bentuk sederhana dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spaceyaitu dengan mengalikan secara distributif dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spacesehingga diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses space end cell equals cell open parentheses square root of 3 cross times square root of 3 close parentheses plus open parentheses square root of 3 cross times 5 close parentheses plus open parentheses negative 2 cross times square root of 3 close parentheses plus open parentheses negative 2 cross times 5 close parentheses end cell row cell open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses end cell equals cell 3 plus 5 square root of 3 minus 2 square root of 3 minus 10 end cell row cell open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses end cell equals cell 3 square root of 3 minus 7 end cell end table end style


Jadi bentuk sederhana dari open parentheses square root of 3 minus 2 close parentheses open parentheses square root of 3 plus 5 close parentheses spaceadalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 3 square root of 3 minus 7 end cell end table

0

Roboguru

Tentukan hasil dari !

Pembahasan Soal:

Ingat kembali rumus berikut.

  • open parentheses a plus b close parentheses open parentheses a minus b close parentheses equals a squared minus b squared 
  • square root of a cross times square root of b equals square root of a cross times b end root 
  • open parentheses square root of a minus square root of b close parentheses squared equals open parentheses square root of a close parentheses squared minus 2 open parentheses square root of a close parentheses open parentheses square root of b close parentheses plus open parentheses square root of b close parentheses squared 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses square root of 3 a plus b end root minus square root of 3 a minus b end root close parentheses squared end cell row blank equals cell open parentheses square root of 3 a plus b end root close parentheses squared minus 2 open parentheses square root of 3 a plus b end root close parentheses open parentheses square root of 3 a minus b end root close parentheses plus open parentheses square root of 3 a minus b end root close parentheses squared end cell row blank equals cell 3 a plus b minus 2 square root of open parentheses 3 a plus b close parentheses open parentheses 3 a minus b close parentheses end root plus 3 a minus b end cell row blank equals cell 3 a up diagonal strike plus b end strike minus 2 square root of 9 a squared minus b squared end root plus 3 a up diagonal strike negative b end strike end cell row blank equals cell 6 a minus 2 square root of 9 a squared minus b squared end root end cell end table  

Jadi,hasil dari open parentheses square root of 3 a plus b end root minus square root of 3 a minus b end root close parentheses squared adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 9 a squared minus b squared end root end cell end table.

0

Roboguru

Tentukan hasil perkalian akar-akar bilangan berikut! e.

Pembahasan Soal:

Ingat kembali rumus berikut.

  • 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 
  • 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 m cross times a to the power of n equals a to the power of m plus n end exponent 
  • square root of a squared cross times b end root equals a square root of b
  • open parentheses square root of a minus square root of b close parentheses open parentheses square root of a plus square root of b close parentheses equals open parentheses square root of a close parentheses squared minus open parentheses square root of b close parentheses squared 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 fourth root of a to the power of 5 end root plus square root of b close parentheses open parentheses 2 fourth root of a to the power of 5 end root minus square root of b close parentheses end cell equals cell 2 squared open parentheses fourth root of a to the power of 5 end root close parentheses squared minus open parentheses square root of b close parentheses squared end cell row blank equals cell 4 open parentheses a to the power of 5 over 4 end exponent close parentheses squared minus b end cell row blank equals cell 4 open parentheses a to the power of 5 over 4 cross times 2 end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a to the power of 5 over 2 end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a to the power of 2 1 half end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a to the power of 2 plus 1 half end exponent close parentheses minus b end cell row blank equals cell 4 open parentheses a squared cross times a to the power of 1 half end exponent close parentheses minus b end cell row blank equals cell 4 a squared square root of a minus b end cell end table 

Jadi, hasil dari open parentheses 2 fourth root of a to the power of 5 end root plus square root of b close parentheses open parentheses 2 fourth root of a to the power of 5 end root minus square root of b close parentheses adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank a end table squared table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell negative b end cell end table.

0

Roboguru

.

Pembahasan Soal:

Gunakan konsep merasionalkan pecahan dengan penyebut yang melibatkan pengurangan bentuk akar serta perkalian bentuk akar.

Akan ditentukan bilangan yang senilai dengan fraction numerator 6 over denominator square root of 7 minus square root of 11 end fraction.

Terlebih dahulu tentukan akar sekawan dari square root of 7 minus square root of 11 yaitu square root of 7 plus square root of 11.

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell fraction numerator 6 over denominator square root of 7 minus square root of 11 end fraction end cell equals cell fraction numerator 6 over denominator square root of 7 minus square root of 11 end fraction cross times fraction numerator square root of 7 plus square root of 11 over denominator square root of 7 plus square root of 11 end fraction end cell row blank equals cell fraction numerator 6 open parentheses square root of 7 plus square root of 11 close parentheses over denominator open parentheses square root of 7 close parentheses squared minus open parentheses square root of 11 close parentheses squared end fraction end cell row blank equals cell fraction numerator 6 open parentheses square root of 7 plus square root of 11 close parentheses over denominator 7 minus 11 end fraction end cell row blank equals cell fraction numerator 6 open parentheses square root of 7 plus square root of 11 close parentheses over denominator negative 4 end fraction end cell row blank equals cell negative 6 over 4 open parentheses square root of 7 plus square root of 11 close parentheses end cell row blank equals cell negative 3 over 2 open parentheses square root of 7 plus square root of 11 close parentheses end cell end table

Jadi, jawaban yang tepat adalah B.

0

Roboguru

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