Roboguru

Tentukan hasil dari: c.

Pertanyaan

Tentukan hasil dari:

c. begin mathsize 14px style 10 square root of 2 plus square root of 12 end style 

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 10 square root of 2 plus square root of 12 end cell equals cell 10 square root of 2 plus square root of 4 cross times 3 end root end cell row blank equals cell 10 square root of 2 plus 2 square root of 3 end cell end table end style 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Narulita

Terakhir diupdate 13 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Jika  dan  maka ...

Pembahasan Soal:

Soal diatas dapat dijawab dengan menyederhanakan masing-masing bentuk akar.

Akan disederhanakan nilai x pada soal sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 1 over 9 plus 1 over 16 end root end cell row blank equals cell square root of fraction numerator 16 plus 9 over denominator 144 end fraction end root end cell row blank equals cell square root of 25 over 144 end root end cell row blank equals cell 5 over 12. end cell end table  

Lalu, nilai y juga akan dioperasikan dan di sederhanakan yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of 1 over 9 end root minus square root of 1 over 16 end root end cell row blank equals cell fraction numerator square root of 1 over denominator square root of 9 end fraction minus fraction numerator square root of 1 over denominator square root of 16 end fraction end cell row blank equals cell 1 third minus 1 fourth end cell row blank equals cell fraction numerator 4 minus 3 over denominator 12 end fraction end cell row blank equals cell 1 over 12. end cell end table  

Sehingga bentuk x dan y paling sederhana adalah x equals 5 over 12 dan y equals 1 over 12.

Dari bentuk yang sederhana tersebut, dapat disimpulkan bahwa nilai x greater than y.


Dengan demikian, jawaban yang benar adalah B.

0

Roboguru

Tentukan hasil operasi akar-akar bilangan berikut! e.

Pembahasan Soal:

Ingat kembali sifat penjumlahan akar berikut!

  • Pada penjumlahan akar dengan bilangan pokok berbeda, maka bilangan pokok harus disamakan terlebih dahulu, kemudian dilanjutkan dengan sifat penjumlahan a cube root of c plus b cube root of c equals left parenthesis a plus b right parenthesis cube root of c   

Dengan rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cube root of 128 plus cube root of 54 end cell equals cell cube root of 64 cross times 2 end root plus cube root of 27 cross times 2 end root end cell row blank equals cell cube root of 4 cubed cross times 2 end root plus cube root of 3 cubed cross times 2 end root end cell row blank equals cell 4 cube root of 2 plus 3 cube root of 2 end cell row blank equals cell left parenthesis 4 plus 3 right parenthesis cube root of 2 end cell row blank equals cell 7 cube root of 2 end cell end table                

Jadi, hasil dari cube root of 128 plus cube root of 54 adalah 7 cube root of 2.

0

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 20 plus square root of 12 cross times 3 square root of 6 end cell equals cell square root of 20 plus 3 square root of 12 cross times 6 end root end cell row blank equals cell square root of 20 plus 3 square root of 72 end cell row blank equals cell square root of 4 cross times 5 end root plus 3 square root of 36 cross times 2 end root end cell row blank equals cell 2 square root of 5 plus 3 cross times 6 square root of 2 end cell row blank equals cell 2 square root of 5 plus 18 square root of 2 end cell end table 

Jadi, square root of 20 plus square root of 12 cross times 3 square root of 6 equals 2 square root of 5 plus 18 square root of 2

0

Roboguru

Hasil dari  adalah ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 12px style 3 open parentheses p to the power of 28 q to the power of negative 12 end exponent close parentheses to the power of 1 fourth end exponent minus 6 cube root of p to the power of 21 q to the power of negative 9 end exponent end root plus fraction numerator square root of cube root of p to the power of 42 end root end root over denominator q cubed end fraction equals 3 open parentheses p to the power of 28 q to the power of negative 12 end exponent close parentheses to the power of 1 fourth end exponent minus 6 open parentheses p to the power of 21 q to the power of negative 9 end exponent close parentheses to the power of 1 third end exponent plus fraction numerator square root of open parentheses p to the power of 42 close parentheses to the power of 1 third end exponent end root over denominator q cubed end fraction blank equals 3 open parentheses p to the power of 28 cross times 1 fourth end exponent q to the power of negative 12 cross times 1 fourth end exponent close parentheses minus 6 open parentheses p to the power of 21 cross times 1 third end exponent q to the power of negative 9 cross times 1 third end exponent close parentheses plus q to the power of negative 3 end exponent square root of open parentheses p to the power of 42 cross times 1 third end exponent close parentheses end root equals 3 p to the power of 7 q to the power of negative 3 end exponent minus 6 p to the power of 7 q to the power of negative 3 end exponent plus q to the power of negative 3 end exponent square root of open parentheses p to the power of 14 close parentheses end root equals 3 p to the power of 7 q to the power of negative 3 end exponent minus 6 p to the power of 7 q to the power of negative 3 end exponent plus q to the power of negative 3 end exponent open parentheses open parentheses p to the power of 14 close parentheses to the power of 1 half end exponent close parentheses equals 3 p to the power of 7 q to the power of negative 3 end exponent minus 6 p to the power of 7 q to the power of negative 3 end exponent plus q to the power of negative 3 end exponent p to the power of 7 equals open parentheses 3 minus 6 plus 1 close parentheses p to the power of 7 q to the power of negative 3 end exponent equals negative 2 p to the power of 7 q to the power of negative 3 end exponent equals negative fraction numerator 2 p to the power of 7 over denominator q cubed end fraction end style 

Jadi, hasil dari undefined adalah undefined.

Dengan demikian, jawaban yang tepat adalah B.

0

Roboguru

Tentukan hasil penjumlahan dan pengurangan dari bentuk akar berikut. c.

Pembahasan Soal:

Perhatikan perhitungan berikut ini!

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 square root of 27 plus square root of 48 minus square root of 12 end cell equals cell 3 square root of 9 cross times 3 end root plus square root of 16 cross times 3 end root minus square root of 4 cross times 3 end root end cell row blank equals cell left parenthesis 3 square root of 9 cross times square root of 3 right parenthesis plus left parenthesis square root of 16 cross times square root of 3 right parenthesis minus left parenthesis square root of 4 cross times square root of 3 right parenthesis end cell row blank equals cell 9 square root of 3 plus 4 square root of 3 minus 2 square root of 3 end cell row blank equals cell square root of 3 open parentheses 9 plus 4 minus 2 close parentheses end cell row blank equals cell square root of 3 open parentheses 11 close parentheses end cell row blank equals cell 11 square root of 3 end cell end table 

Jadi nilai dari begin mathsize 14px style 3 square root of 27 plus square root of 48 minus square root of 12 end style adalah begin mathsize 14px style 11 square root of 3 end style 

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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