Roboguru

Hasil dari 12+108​​32​​−26​48​+3​ adalah ....

Pertanyaan

Hasil dari begin mathsize 14px style fraction numerator square root of 32 over denominator square root of 12 plus square root of 108 end root end fraction minus fraction numerator square root of 48 plus 3 over denominator 2 square root of 6 end fraction end style adalah ....

  1. begin mathsize 14px style fraction numerator 12 square root of 2 minus 11 square root of 6 over denominator 12 end fraction end style 

  2. begin mathsize 14px style fraction numerator 12 square root of 2 minus 5 square root of 6 over denominator 12 end fraction end style 

  3. begin mathsize 14px style fraction numerator 6 square root of 2 minus 11 square root of 6 over denominator 6 end fraction end style 

  4. begin mathsize 14px style fraction numerator 6 square root of 2 minus square root of 6 over denominator 6 end fraction end style 

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 12px style fraction numerator square root of 32 over denominator square root of 12 plus square root of 108 end root end fraction minus fraction numerator square root of 48 plus 3 over denominator 2 square root of 6 end fraction equals fraction numerator square root of 16 times 2 end root over denominator square root of 12 plus square root of 4 times 27 end root end root end fraction minus fraction numerator square root of 16 times 3 end root plus 3 over denominator 2 square root of 6 end fraction equals fraction numerator 4 square root of 2 over denominator square root of 12 plus 2 square root of 27 end root end fraction minus fraction numerator 4 square root of 3 plus 3 over denominator 2 square root of 6 end fraction equals fraction numerator 4 square root of 2 over denominator square root of open parentheses 9 plus 3 close parentheses plus 2 square root of 9 times 3 end root end root end fraction minus fraction numerator 4 square root of 3 plus 3 over denominator 2 square root of 6 end fraction cross times fraction numerator square root of 6 over denominator square root of 6 end fraction equals fraction numerator 4 square root of 2 over denominator square root of 9 plus square root of 3 end fraction minus fraction numerator square root of 6 open parentheses 4 square root of 3 plus 3 close parentheses over denominator 2 times 6 end fraction equals fraction numerator 4 square root of 2 over denominator 3 plus square root of 3 end fraction minus fraction numerator open parentheses square root of 6 cross times 4 square root of 3 close parentheses plus open parentheses square root of 6 cross times 3 close parentheses over denominator 12 end fraction equals fraction numerator 4 square root of 2 open parentheses 3 minus square root of 3 close parentheses over denominator 3 squared minus open parentheses square root of 3 close parentheses squared end fraction minus fraction numerator 4 square root of 18 plus 3 square root of 6 over denominator 12 end fraction equals fraction numerator 4 square root of 2 open parentheses 3 minus square root of 3 close parentheses over denominator 9 minus 3 end fraction minus fraction numerator 12 square root of 2 plus 3 square root of 6 over denominator 12 end fraction equals fraction numerator 12 square root of 2 minus 4 square root of 6 over denominator 6 end fraction minus fraction numerator 12 square root of 2 plus 3 square root of 6 over denominator 12 end fraction equals fraction numerator 24 square root of 2 minus 8 square root of 6 over denominator 12 end fraction minus fraction numerator 12 square root of 2 plus 3 square root of 6 over denominator 12 end fraction equals fraction numerator 24 square root of 2 minus 8 square root of 6 minus 12 square root of 2 minus 3 square root of 6 over denominator 12 end fraction equals fraction numerator 24 square root of 2 minus 12 square root of 2 minus 8 square root of 6 minus 3 square root of 6 over denominator 12 end fraction equals fraction numerator open parentheses 24 minus 12 close parentheses square root of 2 minus open parentheses 8 plus 3 close parentheses square root of 6 over denominator 12 end fraction equals fraction numerator 12 square root of 2 minus 11 square root of 6 over denominator 12 end fraction end style 

Jadi, hasil dari undefined adalah undefined.

Dengan demikian, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Azizatul

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 05 Oktober 2021

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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Pertanyaan yang serupa

Bentuk sederhana dari 11+62​​ 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.

0

Roboguru

Hasil dari 23​5+2​​adalah ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 14px style fraction numerator 5 plus square root of 2 over denominator 2 square root of 3 end fraction equals fraction numerator 5 plus square root of 2 over denominator 2 square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction space space space space space space space space space space space space space equals fraction numerator 5 cross times square root of 3 plus square root of 2 cross times square root of 3 over denominator 2 square root of 3 cross times square root of 3 end fraction space space space space space space space space space space space space space equals fraction numerator 5 square root of 3 plus square root of 6 over denominator 2 cross times 3 end fraction space space space space space space space space space space space space space equals fraction numerator 5 square root of 3 plus square root of 6 over denominator 6 end fraction end style     

Jadi, bentuk rasional dari begin mathsize 14px style fraction numerator 5 plus square root of 2 over denominator 2 square root of 3 end fraction end style adalah undefined.

Dengan demikian, jawaban yang tepat adalah A.

0

Roboguru

Bentuk rasional dari 7​−6​6​+7​​ adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut!

begin mathsize 14px style fraction numerator square root of 6 plus square root of 7 over denominator square root of 7 minus square root of 6 end fraction equals fraction numerator square root of 6 plus square root of 7 over denominator square root of 7 minus square root of 6 end fraction cross times fraction numerator square root of 7 plus square root of 6 over denominator square root of 7 plus square root of 6 end fraction equals fraction numerator left parenthesis square root of 6 cross times square root of 7 right parenthesis plus left parenthesis square root of 7 cross times square root of 7 right parenthesis plus left parenthesis square root of 6 cross times square root of 6 right parenthesis plus left parenthesis square root of 7 cross times square root of 6 right parenthesis over denominator open parentheses square root of 7 close parentheses squared minus open parentheses square root of 6 close parentheses squared end fraction equals fraction numerator square root of 42 plus 7 plus 6 plus square root of 42 over denominator 7 minus 6 end fraction equals fraction numerator 13 plus 2 square root of 42 over denominator 1 end fraction equals 13 plus 2 square root of 42 end style  

Jadi, bentuk rasional dari begin mathsize 14px style fraction numerator square root of 6 plus square root of 7 over denominator square root of 7 minus square root of 6 end fraction end style adalah begin mathsize 14px style 13 plus 2 square root of 42 end style.

Dengan demikian, jawaban yang tepat adalah D.

0

Roboguru

Bentuk sederhana dari 0,34−0,40,3​​ adalah ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 0 , 34 minus 0 , 4 square root of 0 , 3 end root end root end cell equals cell square root of 0 , 34 minus 2 cross times 0 , 2 square root of 0 , 3 end root end root end cell row blank equals cell square root of 0 , 34 minus 2 square root of 0 , 04 cross times 0 , 3 end root end root end cell row blank equals cell square root of open parentheses 0 , 3 plus 0 , 04 close parentheses minus 2 square root of 0 , 3 cross times 0 , 04 end root end root end cell row blank equals cell square root of 0 , 3 end root minus square root of 0 , 04 end root text  (karena 0,3 > 0,04) end text end cell row blank equals cell square root of 0 , 3 end root minus 0 , 2 end cell end table end style 

Dengan demikian, bentuk sederhana dari begin mathsize 14px style square root of 0 , 34 minus 0 , 4 square root of 0 , 3 end root end root end style adalah begin mathsize 14px style square root of 0 , 3 end root minus 0 , 2 end style.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Hasil dari 3​+7​43​​−17−122​​ adalah ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 14px style fraction numerator 4 square root of 3 over denominator square root of 3 plus square root of 7 end fraction minus square root of 17 minus 12 square root of 2 end root equals fraction numerator 4 square root of 3 over denominator square root of 3 plus square root of 7 end fraction cross times fraction numerator square root of 3 minus square root of 7 over denominator square root of 3 minus square root of 7 end fraction minus square root of 17 minus 2 times 6 square root of 2 end root equals fraction numerator 4 square root of 3 open parentheses square root of 3 minus square root of 7 close parentheses over denominator open parentheses square root of 3 close parentheses squared minus open parentheses square root of 7 close parentheses squared end fraction minus square root of 17 minus 2 square root of 72 end root equals fraction numerator open parentheses 4 square root of 3 times square root of 3 close parentheses minus open parentheses 4 square root of 3 times square root of 7 close parentheses over denominator 3 minus 7 end fraction minus square root of open parentheses 9 plus 8 close parentheses minus 2 square root of 9 times 8 end root end root equals fraction numerator 12 minus 4 square root of 21 over denominator negative 4 end fraction minus open parentheses square root of 9 minus square root of 8 close parentheses equals negative fraction numerator 4 open parentheses 3 minus 1 close parentheses square root of 21 over denominator 4 end fraction minus open parentheses 3 minus 2 square root of 2 close parentheses equals negative left parenthesis 3 minus square root of 21 right parenthesis minus 3 plus 2 square root of 2 equals square root of 21 minus 3 minus 3 plus 2 square root of 2 equals square root of 21 plus 2 square root of 2 minus 6 end style 

Jadi, hasil dari undefined adalah undefined.

Dengan demikian, jawaban yang tepat adalah C.

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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