Question

sederhanakan bentuk pangkat berikut!
1.[tex]( \frac{3}{y} )^{ - 1} \times ( \frac{2}{4y}) ^{2} \times ( \frac{5}{3y})^{ - 3} [/tex]
2.
[tex] \frac{(3x)^{2} \div (2x)^{ - 1} \div (2x) ^{ - 3} }{(2x)^{4} \div (3x)^{2} } [/tex]

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Answer 1

Jawab:
[tex]\displaystyle(1.)=\:\bf\frac{9y^2}{500}\\\\\rm(2.)=\:\bf81x^4[/tex]

Penjelasan dengan langkah-langkah:
Sifat eksponen
[tex]\displaystyle\because\:\:\left(\frac{a}{b}\right)^{n}=\frac{a^n}{b^n},\:\:\:\:\left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^{n}=\frac{b^n}{a^n}\:\:\therefore[/tex]
[tex]\displaystyle(1.)\:\:\left(\frac{3}{y}\right)^{-1}\times\left(\frac{2}{4y}\right)^{2}\times\left(\frac{5}{3y}\right)^{-3}\\\\=\:\:\:\:\:\left(\frac{y}{3}\right)^{1}\:\:\:\:\times\left(\frac{2}{4y}\right)^{2}\times\left(\frac{3y}{5}\right)^{3}\\\\=\:\:\:\:\:\frac{y}{3}\:\:\:\:\:\:\:\:\:\:\:\times\frac{2^{2}}{(4y)^{2}}\:\:\:\:\times\frac{(3y)^{3}}{5^3}\\\\\because\:\:\:\:\:\:(ab)^n=a^nb^n\:\:\:\:\:\:\therefore[/tex]

[tex]\displaystyle=\:\:\:\:\:\frac{y}{3}\:\:\:\:\:\:\:\:\:\:\times\frac{2^{2}}{(4^{2})y^{2}}\:\:\:\:\times\frac{3^{3}y^{3}}{5^3}\\\\=\:\:\:\:\:\frac{y}{3}\:\:\:\:\:\:\:\:\:\:\times\frac{4}{16y^{2}}\:\:\:\:\:\:\:\:\times\frac{27y^{3}}{125}\\\\=\:\:\:\:\:\frac{y}{3}\:\:\:\:\:\:\:\:\:\:\times\:\:\frac{1}{4y^{2}}\:\:\:\:\:\:\:\:\times\frac{27y^{3}}{125}\\\\=\:\:\:\:\:\frac{27y}{3}\:\:\:\:\:\:\times\:\:\frac{1}{4y^{2}}\:\:\:\:\:\:\:\:\times\frac{y^{3}}{125}[/tex]
[tex]\displaystyle=\:\:\:\:\:9y\:\:\:\:\:\:\:\:\:\times\:\:\frac{1}{4y^{2}}\:\:\:\:\:\:\:\:\times\frac{y^{3}}{125}\\\\=\:\:\:\:\:\frac{9y(y^3)}{4(125)y^2}\:\:\:\:\:\:\:\:=\:\:\:\:\:\frac{9y(y^3)}{500y^2}\\\\=\:\:\:\:\:\frac{9}{500}\:\:\:\:\:\:\times\frac{y^1(y^3)}{y^2}\\\\\because n^a(n^b)=n^{a+b},\:\:\:\frac{n^a}{n^b}=n^{a-b}\therefore\\\\=\:\:\:\:\:\frac{9}{500}\:\:\:\:\:\:\times y^{1+3-2}\\\\=\:\:\:\:\:\frac{9}{500}\:\:\:\:\:\:\times y^{2}\\\\=\:\:\:\:\:\bf\frac{9y^2}{500}[/tex]

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[tex]\displaystyle(2.)\:\:\:\frac{(3x)^2\div(2x)^{-1}\div(2x)^{-3}}{(2x)^4\div(3x)^2}\\\\=\:\:\:\:\:\frac{(3x)^2\div((2x)^{-1}\times(2x)^{-3})}{(2x)^4\div(3x)^2}\\\\\\=\:\:\:\:\:(3x)^2\times(3x)^2\div((2x)^{-1}\times(2x)^{-3}\times(2x)^4)\\\\=\:\:\:\:\:((3x)^2)^2\:\:\:\:\:\:\:\:\:\:\div((2x)^{-1}\times(2x)^{-3}\times(2x)^4)\\\\\because\:\:\:\:\:(ab)^n=a^n(b^n),\:\:\:\:\:n^a\times n^b=n^{a+b}\:\:\:\:\:\therefore\\\\=\:\:\:\:\:((3^2)x^2)^2\:\:\:\:\:\:\:\:\:\:\div((2x)^{-1+(-3)+4})[/tex]

[tex]\displaystyle=\:\:\:\:\:(3^2)^2(x^2)^2\:\:\:\:\:\:\:\:\div((2x)^{0})\\\\=\:\:\:\:\:9^2(x^4)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\div\not1\\\\=\:\:\:\:\:\bf81x^4[/tex]

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(xcvi)