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Bentuk sederhana dari square root of 20 plus square root of 20 plus square root of 20 plus horizontal ellipsis end root end root end root equals horizontal ellipsis 

  1. 3

  2. 4

  3. 5

  4. 6

  5. 6,5

Pembahasan Video:

Pembahasan Soal:

Misalkan square root of 20 plus square root of 20 plus square root of 20 plus horizontal ellipsis end root end root end root equals a, maka dapat dituliskan menjadi:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 20 plus square root of 20 plus square root of 20 plus horizontal ellipsis end root end root end root end cell equals a row cell square root of 20 plus a end root end cell equals cell a space open parentheses kuadratkan space kedua space ruas close parentheses end cell row cell 20 plus a end cell equals cell a squared end cell row cell a squared minus a minus 20 end cell equals 0 row cell open parentheses a minus 5 close parentheses open parentheses a plus 4 close parentheses end cell equals 0 row a equals cell 5 space atau space a equals negative 4 end cell end table 

Karena nilai akar tidak mungkin negatif, jadi bentuk sederhana dari square root of 20 plus square root of 20 plus square root of 20 plus horizontal ellipsis end root end root end root equals 5.

Jadi, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 13 September 2021

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Pembahasan Soal:

Berdasarkan sifat bilangan akar yaitu square root of open parentheses a plus b close parentheses plus-or-minus 2 square root of a. b end root end root equals square root of a plus-or-minus square root of b maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 over denominator square root of open parentheses 2 plus 3 close parentheses plus 2 square root of 2 cross times 3 end root end root end fraction end cell row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 over denominator square root of 2 plus square root of 3 end fraction end cell end table


Ketika ada penyebut berbentuk akar maka perlu dilakukan rasionalisasi bentuk akar dengan mengalikan akar sekawanannya sehingga diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 over denominator square root of 2 plus square root of 3 end fraction end cell row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 over denominator square root of 2 plus square root of 3 end fraction cross times fraction numerator square root of 2 minus square root of 3 over denominator square root of 2 minus square root of 3 end fraction end cell row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 open parentheses square root of 2 minus square root of 3 close parentheses over denominator open parentheses square root of 2 cross times square root of 2 close parentheses up diagonal strike negative open parentheses square root of 2 cross times square root of 3 close parentheses plus open parentheses square root of 2 cross times square root of 3 close parentheses end strike minus open parentheses square root of 3 cross times square root of 3 close parentheses end fraction end cell row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 open parentheses square root of 2 minus square root of 3 close parentheses over denominator 2 minus 3 end fraction end cell row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell fraction numerator 4 open parentheses square root of 2 minus square root of 3 close parentheses over denominator negative 1 end fraction end cell row cell fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction end cell equals cell 4 square root of 3 minus 4 square root of 2 end cell end table end style


Jadi nilai dari fraction numerator 4 over denominator square root of 5 plus 2 square root of 6 end root end fraction adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 4 square root of 3 minus 4 square root of 2. end cell end table

Roboguru

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut ini!

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 11 plus 6 square root of 2 end root end cell equals cell square root of 11 plus 2 cross times 3 square root of 2 end root end cell row blank equals cell square root of 11 plus 2 square root of 9 cross times 2 end root end root end cell row blank equals cell square root of left parenthesis 9 plus 2 right parenthesis plus 2 square root of 9 cross times 2 end root end root end cell end table 

Ingat bahwa square root of left parenthesis x plus y right parenthesis plus 2 square root of x y end root end root equals square root of x plus square root of y, maka dari perolehan diatas, akan didapatkan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 11 plus 6 square root of 2 end root end cell equals cell square root of left parenthesis 9 plus 2 right parenthesis plus 2 square root of 9 cross times 2 end root end root end cell row blank equals cell square root of 9 plus square root of 2 end cell row blank equals cell 3 plus square root of 2 end cell end table

Dengan demikian, bentuk sederhana dari begin mathsize 14px style square root of 11 plus 6 square root of 2 end root end style adalah begin mathsize 14px style 3 plus square root of 2 end style.

Jadi, jawaban yang tepat adalah A.

Roboguru

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut!

begin mathsize 14px style square root of 0 , 41 plus 0 , 2 square root of 0 , 4 end root end root equals square root of 0 , 41 plus 2 times 0 , 1 square root of 0 , 4 end root end root equals square root of 0 , 41 plus 2 square root of open parentheses 0 , 1 close parentheses squared times 0 , 4 end root end root equals square root of 0 , 41 plus 2 square root of 0 , 01 times 0 , 4 end root end root equals square root of left parenthesis 0 , 4 plus 0 , 01 right parenthesis plus 2 square root of 0 , 4 times 0 , 01 end root end root space end style

Karena begin mathsize 14px style square root of open parentheses a plus b close parentheses plus 2 square root of a b end root end root equals square root of a plus square root of b end style, maka didapat perhitungan sebagai berikut.

begin mathsize 14px style square root of left parenthesis 0 , 4 plus 0 , 01 right parenthesis plus 2 square root of 0 , 4 times 0 , 01 end root end root equals square root of 0 , 4 end root plus square root of 0 , 01 end root equals square root of 0 , 1 times 4 end root plus 0 , 1 equals 2 square root of 0 , 1 end root plus 0 , 1 end style

Dengan demikian, bentuk sederhana dari begin mathsize 14px style square root of 0 , 41 plus 0 , 2 square root of 0 , 4 end root end root end style adalah begin mathsize 14px style 2 square root of 0 , 1 end root plus 0 , 1 end style.

Jadi, jawaban yang tepat adalah B.

Roboguru

Bentuk sederhana dari  adalah ...

Pembahasan Soal:

fraction numerator 10 over denominator 4 minus square root of 11 end fraction equals fraction numerator 10 over denominator 4 minus square root of 11 end fraction cross times fraction numerator 4 plus square root of 11 over denominator 4 plus square root of 11 end fraction space space space space space space space space space space space space space space space equals fraction numerator 10 open parentheses 4 plus square root of 11 close parentheses over denominator 16 minus 11 end fraction space space space space space space space space space space space space space space space equals fraction numerator 10 open parentheses 4 plus square root of 11 close parentheses over denominator 5 end fraction space space space space space space space space space space space space space space space equals 2 open parentheses 4 plus square root of 11 close parentheses space space space space space space space space space space space space space space space equals 8 plus 2 square root of 11 

Jadi, bentuk sederhana dari fraction numerator bold 10 over denominator bold 4 bold minus square root of bold 11 end fraction adalah bold 8 bold plus bold 2 square root of bold 11

Oleh karena itu, jawaban yang tepat adalah A.

Roboguru

6. Bentuk sederhana dari  adalah...

Pembahasan Soal:

Perhatikan penghitungan berikut!

 fraction numerator 3 square root of 2 over denominator 2 square root of 3 minus square root of 11 end fraction equals fraction numerator 3 square root of 2 over denominator 2 square root of 3 minus square root of 11 end fraction cross times fraction numerator 2 square root of 3 plus square root of 11 over denominator 2 square root of 3 plus square root of 11 end fraction space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 square root of 2 space open parentheses 2 square root of 3 plus square root of 11 close parentheses over denominator open parentheses 2 square root of 3 close parentheses squared minus open parentheses square root of 11 close parentheses squared end fraction space space space space space space space space space space space space space space space space space space space space equals fraction numerator 3 square root of 2 space open parentheses 2 square root of 3 plus square root of 11 close parentheses over denominator 12 minus 11 end fraction space space space space space space space space space space space space space space space space space space space space equals 3 square root of 2 space open parentheses 2 square root of 3 plus square root of 11 close parentheses space space space space space space space space space space space space space space space space space space space space equals 3 space open parentheses 2 square root of 2 square root of 3 plus square root of 2 square root of 11 close parentheses space space space space space space space space space space space space space space space space space space space space equals 3 space open parentheses 2 square root of 6 plus square root of 22 close parentheses 

Oleh karena itu, jawaban yang benar adalah D. undefined 

 

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