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Bentuk sederhana dari

Pertanyaan

Bentuk sederhana dari fraction numerator x to the power of negative 2 end exponent plus y to the power of negative 3 end exponent over denominator x to the power of begin display style 1 half end style end exponent minus y to the power of begin display style 1 half end style end exponent end fraction

Pembahasan Soal:

Bentuk sederhana dari fraction numerator x to the power of negative 2 end exponent plus y to the power of negative 3 end exponent over denominator x to the power of begin display style 1 half end style end exponent minus y to the power of begin display style 1 half end style end exponent end fraction adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x to the power of negative 2 end exponent plus y to the power of negative 3 end exponent over denominator x to the power of begin display style 1 half end style end exponent minus y to the power of begin display style 1 half end style end exponent end fraction end cell equals cell fraction numerator begin display style 1 over x squared end style plus begin display style 1 over y cubed end style over denominator square root of x minus square root of y end fraction end cell end table

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Umi

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of open parentheses fraction numerator p squared minus q squared over denominator p plus q end fraction close parentheses squared end root square root of fraction numerator p plus q over denominator p squared minus q squared end fraction end root end cell equals cell open parentheses fraction numerator p squared minus q squared over denominator p plus q end fraction close parentheses squared open parentheses fraction numerator p plus q over denominator p squared minus q squared end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator up diagonal strike open parentheses p plus q close parentheses end strike open parentheses p minus q close parentheses over denominator up diagonal strike open parentheses p plus q close parentheses end strike end fraction close parentheses squared open parentheses fraction numerator up diagonal strike open parentheses p plus q close parentheses end strike over denominator up diagonal strike open parentheses p plus q close parentheses end strike open parentheses p minus q close parentheses end fraction close parentheses end cell row blank equals cell open parentheses p minus q close parentheses to the power of up diagonal strike 2 end exponent fraction numerator 1 over denominator up diagonal strike open parentheses p minus q close parentheses end strike end fraction end cell row blank equals cell p minus q end cell end table

Jadi, jawaban yang tepat adalah E

3

Roboguru

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 plus square root of 8 over denominator square root of 6 end fraction end cell equals cell fraction numerator 2 over denominator square root of 6 end fraction plus fraction numerator square root of 8 over denominator square root of 6 end fraction end cell row blank equals cell open parentheses fraction numerator 2 over denominator square root of 6 end fraction cross times fraction numerator square root of 6 over denominator square root of 6 end fraction close parentheses plus open parentheses fraction numerator square root of 8 over denominator square root of 6 end fraction cross times fraction numerator square root of 6 over denominator square root of 6 end fraction close parentheses end cell row blank equals cell 2 over 6 square root of 6 plus fraction numerator square root of 48 over denominator 6 end fraction end cell row blank equals cell 2 over 6 square root of 6 plus fraction numerator square root of 16 cross times 3 end root over denominator 6 end fraction end cell row blank equals cell 2 over 6 square root of 6 plus 4 over 6 square root of 3 end cell row blank equals cell 1 third square root of 6 plus 2 over 3 square root of 3 end cell end table 

Oleh karena itu, jawaban yang benar adalah C.

2

Roboguru

Hasil perkalian  adalah

Pembahasan Soal:

Ingat sifat distributif perkalian pada pengurangan yaitu 

a open parentheses b minus c close parentheses equals a b minus a c

sehingga square root of 6 open parentheses 3 square root of 2 minus square root of 32 close parentheses diperoleh sebagai berikut.

square root of 6 open parentheses 3 square root of 2 minus square root of 32 close parentheses equals square root of 6 cross times 3 square root of 2 minus square root of 6 square root of 32

Berdasarkan aturan perkalian dengan akar pangkat sama yaitu

n-th root of a cross times n-th root of b equals n-th root of a cross times b end root comma space a not equal to 0 text  dan end text space b not equal to 0

Sehingga 

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 6 open parentheses 3 square root of 2 minus square root of 32 close parentheses end cell equals cell square root of 6 cross times 3 square root of 2 minus square root of 6 square root of 32 end cell row blank equals cell 3 square root of 6 cross times 2 end root minus square root of 6 cross times 32 end root end cell row blank equals cell 3 square root of 12 minus square root of 192 end cell end table

Jika n-th root of a dapat difaktorkan menjadi n-th root of p cross times q end root dengan p space text atau end text space q bilangan berpangkat n maka n-th root of a dapat disederhanakan sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 6 open parentheses 3 square root of 2 minus square root of 32 close parentheses end cell equals cell 3 square root of 12 minus square root of 192 end cell row blank equals cell 3 square root of 4 cross times 3 end root minus square root of 64 cross times 3 end root end cell row blank equals cell 3 square root of 2 squared cross times 3 end root minus square root of 8 squared cross times 3 end root end cell row blank equals cell 3 cross times 2 square root of 3 minus 8 square root of 3 end cell row blank equals cell 6 square root of 3 minus 8 square root of 3 end cell end table

Jika p n-th root of a minus q n-th root of a equals open parentheses p minus q close parentheses n-th root of a maka

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 6 open parentheses 3 square root of 2 minus square root of 32 close parentheses end cell equals cell 6 square root of 3 minus 8 square root of 3 end cell row blank equals cell open parentheses 6 minus 8 close parentheses square root of 3 end cell row blank equals cell negative 2 square root of 3 end cell end table

Jadi hasil perkalian square root of 6 open parentheses 3 square root of 2 minus square root of 32 close parentheses adalah negative 2 square root of 3.

Oleh karena itu, jawaban yang benar adalah A.

1

Roboguru

Tentukan hasil pemangkatan akar-akar bilangan berikut! b.

Pembahasan Soal:

Ingat kembali rumus berikut.

square root of a cross times square root of b equals square root of a cross times b end root 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 10 close parentheses to the power of 4 end cell equals cell square root of 10 cross times square root of 10 cross times square root of 10 cross times square root of 10 end cell row blank equals cell square root of 10 cross times 10 end root cross times square root of 10 cross times 10 end root end cell row blank equals cell square root of 10 squared end root cross times square root of 10 squared end root end cell row blank equals cell 10 cross times 10 end cell row blank equals 100 end table 

Jadi, hasil dari open parentheses square root of 10 close parentheses to the power of 4 adalah 100.

0

Roboguru

Tentukan hasil perkalian akar-akar bilangan berikut! f.

Pembahasan Soal:

Ingat kembali sifat perkalian akar berikut!

Jika a space dan space b sebarang bilangan bulat, maka berlaku:

square root of a cross times square root of b equals square root of a cross times b end root 

Dengan rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 square root of 21 cross times 5 square root of 3 end cell equals cell 2 cross times square root of 21 cross times 5 cross times square root of 3 end cell row blank equals cell 2 cross times 5 cross times square root of 21 cross times square root of 3 end cell row blank equals cell 10 cross times square root of 21 cross times 3 end root end cell row blank equals cell 10 cross times square root of 63 end cell row blank equals cell 10 cross times square root of 9 cross times 7 end root end cell row blank equals cell 10 cross times 3 square root of 7 end cell row blank equals cell 30 square root of 7 end cell end table    

Jadi, hasil dari 2 square root of 21 cross times 5 square root of 3 equals 30 square root of 7.

2

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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