Roboguru

Bentuk sederhana dari:   adalah ...

Pertanyaan

Bentuk sederhana dari:

fraction numerator 2 over denominator square root of 12 minus square root of 8 end fraction 

adalah ...begin mathsize 14px style space end style 

  1. square root of 3 plus square root of 2 

  2. 2 square root of 3 minus square root of 2 

  3. square root of 3 minus 1 fifth square root of 2 

  4. 2 open parentheses square root of 3 minus square root of 2 close parentheses 

  5. 2 over 4 square root of 12 minus square root of 8 

Pembahasan Soal:

Merasionalkan bentuk akar.

fraction numerator a over denominator square root of b minus square root of c end fraction equals fraction numerator a over denominator square root of b minus square root of c end fraction times fraction numerator square root of b plus square root of c over denominator square root of b plus square root of c end fraction 

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 over denominator square root of 12 minus square root of 8 end fraction end cell equals cell fraction numerator 2 over denominator square root of 12 minus square root of 8 end fraction times fraction numerator square root of 12 plus square root of 8 over denominator square root of 12 plus square root of 8 end fraction end cell row blank equals cell fraction numerator 2 open parentheses square root of 12 plus square root of 8 close parentheses over denominator 12 minus 8 end fraction end cell row blank equals cell fraction numerator 2 open parentheses square root of 4 times 3 end root minus square root of 4 times 2 end root close parentheses over denominator 4 end fraction end cell row blank equals cell fraction numerator 2 open parentheses 2 square root of 3 minus 2 square root of 2 close parentheses over denominator 4 end fraction end cell row blank equals cell fraction numerator 4 square root of 3 minus 4 square root of 2 over denominator 4 end fraction end cell row blank equals cell square root of 3 minus square root of 2 end cell end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 over denominator square root of 12 minus square root of 8 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 3 end cell 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 cell square root of 2 end cell end table.

Jadi, tidak ada jawaban yang tepat.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Indraprasta PGRI

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Bentuk sederhana dari permasalahan tersebut dapat diperoleh dengan mengalikan akar sekawan.

Perhatikan perhitungan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell fraction numerator 2 plus 3 square root of 2 over denominator 2 minus 2 square root of 2 end fraction end cell equals cell fraction numerator 2 plus 3 square root of 2 over denominator 2 minus 2 square root of 2 end fraction cross times fraction numerator 2 plus 2 square root of 2 over denominator 2 plus 2 square root of 2 end fraction end cell row blank equals cell fraction numerator open parentheses 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 square root of 2 close parentheses plus open parentheses 3 square root of 2 cross times 2 close parentheses plus open parentheses 3 square root of 2 cross times 2 square root of 2 close parentheses over denominator 2 squared minus open parentheses 2 square root of 2 close parentheses squared end fraction end cell row blank equals cell fraction numerator 4 plus 4 square root of 2 plus 6 square root of 2 plus 12 over denominator 4 minus 8 end fraction end cell row blank equals cell fraction numerator 16 plus 10 square root of 2 over denominator negative 4 end fraction end cell row cell fraction numerator 2 plus 3 square root of 2 over denominator 2 minus 2 square root of 2 end fraction end cell equals cell negative 1 fourth open parentheses 10 square root of 2 plus 16 close parentheses end cell end table end style 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Bentuk sederhana dari

Pembahasan Soal:

Dengan cara merasionalkan bentuk akar (Mengalikan akar sekawan) maka :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 12 over denominator square root of 6 plus square root of 2 end fraction end cell equals cell fraction numerator 12 over denominator square root of 6 plus square root of 2 end fraction cross times fraction numerator square root of 6 minus square root of 2 over denominator square root of 6 minus square root of 2 end fraction end cell row blank equals cell fraction numerator 12 open parentheses square root of 6 minus square root of 2 close parentheses over denominator 6 minus 2 end fraction end cell row blank equals cell fraction numerator 12 left parenthesis square root of 6 minus square root of 2 right parenthesis over denominator 4 end fraction end cell row blank equals cell 3 left parenthesis square root of 6 minus square root of 2 right parenthesis end cell row blank equals cell 3 square root of 6 minus 3 square root of 2 end cell end table end style

Jadi, bentuk sederhada dari begin mathsize 14px style fraction numerator 12 over denominator square root of 6 plus square root of 2 end fraction end style adalah begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 6 end cell 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 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table end style

1

Roboguru

Bentuk rasional penyebut dari pecahan :  adalah ....

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 over denominator 5 minus square root of 13 end fraction end cell equals cell fraction numerator 2 over denominator 5 minus square root of 13 end fraction cross times fraction numerator 5 plus square root of 13 over denominator 5 plus square root of 13 end fraction end cell row blank equals cell fraction numerator 2 open parentheses 5 plus square root of 13 close parentheses over denominator 25 minus 13 end fraction end cell row blank equals cell fraction numerator 2 open parentheses 5 plus square root of 13 close parentheses over denominator 12 end fraction end cell row blank equals cell fraction numerator 5 plus square root of 13 over denominator 6 end fraction end cell end table 

Jadi, bentuk rasionalnya adalah fraction numerator 5 plus square root of 13 over denominator 6 end fraction.

0

Roboguru

Bentuk rasional dari  adalah ....

Pembahasan Soal:

Ingat:

begin mathsize 14px style fraction numerator 1 over denominator square root of a end fraction equals fraction numerator 1 over denominator square root of a end fraction cross times fraction numerator square root of a over denominator square root of a end fraction equals fraction numerator square root of a over denominator a end fraction end style

maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator square root of 12 minus square root of 6 over denominator square root of 3 end fraction end cell equals cell fraction numerator square root of 12 minus square root of 6 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell fraction numerator open parentheses square root of 12 close parentheses open parentheses square root of 3 close parentheses minus open parentheses square root of 6 close parentheses open parentheses square root of 3 close parentheses over denominator open parentheses square root of 3 close parentheses open parentheses square root of 3 close parentheses end fraction end cell row blank equals cell fraction numerator square root of 36 minus square root of 18 over denominator 3 end fraction end cell row blank equals cell fraction numerator 6 minus 3 square root of 2 over denominator 3 end fraction end cell row blank equals cell 2 minus square root of 2 end cell end table end style

Jadi, jawaban yang tepat adalah A.

1

Roboguru

Bentuk rasional dari  adalah ...

Pembahasan Soal:

Untuk merasionalkan begin mathsize 14px style fraction numerator 8 over denominator 3 minus square root of 5 end fraction end style, dapat dilakukan dengan mengalikan pembilang dan penyebut dengan sekawan dari begin mathsize 14px style 3 minus square root of 5 end style yaitu  begin mathsize 14px style 3 plus square root of 5 end style. Diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 8 over denominator 3 minus square root of 5 end fraction cross times fraction numerator 3 plus square root of 5 over denominator 3 plus square root of 5 end fraction end cell equals cell fraction numerator 8 open parentheses 3 plus square root of 5 close parentheses over denominator open parentheses 3 minus square root of 5 close parentheses open parentheses 3 plus square root of 5 close parentheses end fraction end cell row blank equals cell fraction numerator 24 plus 8 square root of 5 over denominator 9 minus 5 end fraction end cell row blank equals cell fraction numerator 24 plus 8 square root of 5 over denominator 4 end fraction end cell row blank equals cell 6 plus 2 square root of 5 end cell end table end style  


sehingga bentuk rasional dari begin mathsize 14px style fraction numerator 8 over denominator 3 minus square root of 5 end fraction end style adalah begin mathsize 14px style 6 plus 2 square root of 5 end style.

Jadi, jawaban yang benar adalah D.

0

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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