Sederhanakan bentuk-bentuk di bawah ini dengan menggunakan sifat-sifat bilangan pangkat! f. (−6u3v)4÷(2uv2)2

Pertanyaan

Sederhanakan bentuk-bentuk di bawah ini dengan menggunakan sifat-sifat bilangan pangkat!

f. $\left(-6u^3v\right)^4\div\left(2uv^2\right)^2$

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

$a to the power of n equals stack a cross times a cross times a cross times horizontal ellipsis cross times a with underbrace below table row blank cell space space space space space space space space end cell cell n space text faktor end text end cell end table$

Ingat sifat bilangan berpangkat berikut.

$a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent$

$open parentheses a b close parentheses to the power of m equals a to the power of m times b to the power of m$

$open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent$

$a to the power of 0 equals 1 comma space a not equal to 0$

sehingga bentuk sederhana dari soal di atas, yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 6 u cubed v close parentheses to the power of 4 divided by open parentheses 2 u v squared close parentheses squared end cell row blank equals cell open parentheses open parentheses negative 6 close parentheses to the power of 4 open parentheses u cubed close parentheses to the power of 4 open parentheses v close parentheses to the power of 4 close parentheses divided by open parentheses open parentheses 2 close parentheses squared open parentheses u close parentheses squared open parentheses v squared close parentheses squared close parentheses end cell row blank equals cell open parentheses open parentheses 2 cross times open parentheses negative 3 close parentheses close parentheses to the power of 4 times u to the power of 12 times v to the power of 4 close parentheses divided by open parentheses 2 squared times u squared times v to the power of 4 close parentheses end cell row blank equals cell open parentheses 2 to the power of 4 times open parentheses negative 3 close parentheses to the power of 4 times u to the power of 12 times v to the power of 4 close parentheses divided by open parentheses 2 squared times u squared times v to the power of 4 close parentheses end cell row blank equals cell 2 to the power of 4 minus 2 end exponent times open parentheses negative 3 close parentheses to the power of 4 times u to the power of 12 minus 2 end exponent times v to the power of 4 minus 4 end exponent end cell row blank equals cell 2 squared times 81 times u to the power of 10 times v to the power of 0 end cell row blank equals cell 2 squared times 3 to the power of 4 times u to the power of 10 times 1 end cell row blank equals cell open parentheses 2 cross times 3 squared close parentheses squared times u to the power of 10 end cell row blank equals cell 18 squared times u to the power of 10 end cell end table$

Dengan demikian, bentuk sederhana dari $table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 6 u cubed v close parentheses to the power of 4 divided by open parentheses 2 u v squared close parentheses squared end cell equals cell 18 squared times u to the power of 10 end cell end table$

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 07 Oktober 2021

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Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

Ingat sifat bilangan berpangkat berikut.

$open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent$

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-Bilangan pokok negatif berpangkat genap hasilnya bilangan positif

sehingga bentuk sederhana dari soal di atas, yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative k cubed close parentheses squared divided by k to the power of 4 end cell equals cell open parentheses negative k close parentheses to the power of 3 times 2 end exponent divided by k to the power of 4 end cell row blank equals cell open parentheses negative k close parentheses to the power of 6 divided by k to the power of 4 end cell row blank equals cell k to the power of 6 divided by k to the power of 4 end cell row blank equals cell k to the power of 6 minus 4 end exponent end cell row blank equals cell k squared end cell end table$

Dengan demikian, bentuk sederhana dari $table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative k cubed close parentheses squared divided by k to the power of 4 end cell equals cell k squared end cell end table$

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Sederhanakan bentuk-bentuk di bawah ini dengan menggunakan sifat-sifat bilangan pangkat! d. (−5m5÷m2)4×m7

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

Ingat sifat bilangan berpangkat berikut.

$a to the power of m times a to the power of n equals a to the power of m plus n end exponent$

$a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent$

$open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent$

sehingga bentuk sederhana dari soal di atas, yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative m to the power of 5 divided by m squared close parentheses to the power of 4 cross times m to the power of 7 end cell row blank equals cell open parentheses negative open parentheses m to the power of 5 minus 2 end exponent close parentheses close parentheses to the power of 4 cross times m to the power of 7 end cell row blank equals cell open parentheses negative m cubed close parentheses to the power of 4 cross times m to the power of 7 end cell row blank equals cell open parentheses negative m close parentheses to the power of 3 times 4 end exponent cross times m to the power of 7 end cell row blank equals cell open parentheses negative m close parentheses to the power of 12 cross times m to the power of 7 end cell row blank equals cell m to the power of 12 cross times m to the power of 7 end cell row blank equals cell m to the power of 12 plus 7 end exponent end cell row blank equals cell m to the power of 19 end cell end table$

Dengan demikian, bentuk sederhana dari $table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative m to the power of 5 divided by m squared close parentheses to the power of 4 cross times m to the power of 7 end cell equals cell m to the power of 19 end cell end table$

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Hasil dari (−3)3+(−3)2+(−3)1+(−3)0 adalah ...

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

Ingat sifat bilangan berpangkat berikut.

$a to the power of 1 equals a$

$a to the power of 0 equals 1 comma space a not equal to 0$

sehingga penyelesaian soal di atas, yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 3 close parentheses cubed plus open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses to the power of 1 plus open parentheses negative 3 close parentheses to the power of 0 end cell row blank equals cell open parentheses negative 3 close parentheses open parentheses negative 3 close parentheses open parentheses negative 3 close parentheses plus open parentheses negative 3 close parentheses open parentheses negative 3 close parentheses plus open parentheses negative 3 close parentheses plus 1 end cell row blank equals cell negative 27 plus 9 plus open parentheses negative 3 close parentheses plus 1 end cell row blank equals cell negative 20 end cell end table$

Dengan demikian, hasil dari $table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 3 close parentheses cubed plus open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses to the power of 1 plus open parentheses negative 3 close parentheses to the power of 0 end cell equals cell negative 20 end cell end table$

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Hasil pemangkatan (4b3a2)−3 adalah ...

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

Ingat sifat bilangan berpangkat berikut.

$a to the power of negative m end exponent equals 1 over a to the power of m comma space a not equal to 0$

$open parentheses a over b close parentheses to the power of m equals a to the power of m over b to the power of m comma space b not equal to 0$

$open parentheses a b close parentheses to the power of m equals a to the power of m times b to the power of m$

$open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent$

sehingga penyelesaian dari soal di atas, yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator 3 a squared over denominator 4 b end fraction close parentheses to the power of negative 3 end exponent end cell equals cell open parentheses 3 a squared close parentheses to the power of negative 3 end exponent over open parentheses 4 b close parentheses to the power of negative 3 end exponent end cell row blank equals cell open parentheses 4 b close parentheses cubed over open parentheses 3 a squared close parentheses cubed end cell row blank equals cell fraction numerator 4 cubed b cubed over denominator 3 cubed a to the power of 6 end fraction end cell row blank equals cell fraction numerator 64 b cubed over denominator 27 a to the power of 6 end fraction end cell end table$

Oleh karena itu, jawaban yang tepat adalah D.

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Gunakan sifat-sifat operasi bilangan berpangkat untuk menyederhanakan operasi aljabar berikut. d. (3p2q3)4(32p3q4)3

Pembahasan Soal:

Bilangan berpangkat bulat positif dapat didefinisikan sebagai berikut.

Ingat sifat bilangan berpangkat berikut.

$a to the power of m divided by a to the power of n equals a to the power of m minus n end exponent$

$open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent$

$open parentheses a b close parentheses to the power of m equals a to the power of m b to the power of m$

$a to the power of 0 equals 1 comma space a not equal to 0$

Penyelesaian soal di atas, yaitu

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 squared p cubed q to the power of 4 close parentheses cubed over open parentheses 3 p squared q cubed close parentheses to the power of 4 end cell equals cell fraction numerator open parentheses 3 squared close parentheses cubed open parentheses p cubed close parentheses cubed open parentheses q to the power of 4 close parentheses cubed over denominator 3 to the power of 4 open parentheses p squared close parentheses to the power of 4 open parentheses q cubed close parentheses to the power of 4 end fraction end cell row blank equals cell fraction numerator 3 to the power of 6 p to the power of 9 q to the power of 12 over denominator 3 to the power of 4 p to the power of 8 q to the power of 12 end fraction end cell row blank equals cell 3 to the power of 6 minus 4 end exponent p to the power of 9 minus 8 end exponent q to the power of 12 minus 12 end exponent end cell row blank equals cell 3 squared p q to the power of 0 end cell row blank equals cell 3 squared p times 1 end cell row blank equals cell 9 p end cell end table$

Dengan demikian, bentuk sederhana dari $table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 squared p cubed q to the power of 4 close parentheses cubed over open parentheses 3 p squared q cubed close parentheses to the power of 4 end cell equals cell 9 p end cell end table$

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