Exercise Zone : Pangkat (Eksponen)

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No. 1

Hitunglah hasil perpangkatan berikut $${\left({\displaystyle \frac{27}{125}}\right)}^{\frac{1}{3}}{\left({\displaystyle \frac{1}{9}}\right)}^{-2}$$

Matematika Idhamdaz

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\left({\displaystyle \frac{27}{125}}\right)}^{\frac{1}{3}}{\left({\displaystyle \frac{1}{9}}\right)}^{-2}& ={\left({\displaystyle \frac{3^3}{5^3}}\right)}^{\frac{1}{3}}{\left({\displaystyle \frac{1}{3^2}}\right)}^{-2}\\ & ={\displaystyle \frac{{\left(3^3\right)}^{\frac{1}{3}}}{{\left(5^3\right)}^{\frac{1}{3}}}}{\left(3^{-2}\right)}^{-2}\\ & ={\displaystyle \frac{3}{5}}\left(3^4\right)\\ & ={\displaystyle \frac{3}{5}}\left(81\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{243}{5}}}}}}\end{array}$$

Jadi, \left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}=\dfrac{243}5.

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No. 2

Diketahui a, b, dan c adalah bilangan real positif, jika {\dfrac{\sqrt{bc}}{\sqrt[5]{ab^4}}= ab}, maka nilai c⁵ adalah

1. a¹¹b¹²
2. a¹²b¹¹
3. a¹⁴b¹³

4. a¹³b¹²
5. a¹²b¹³

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{\sqrt{bc}}{\sqrt[5]{a{b}^4}}}& =ab\\ {\left({\displaystyle \frac{\sqrt{bc}}{\sqrt[5]{a{b}^4}}}\right)}^{10}& =(ab{)}^{10}\\ {\displaystyle \frac{(bc{)}^5}{{\left(a{b}^4\right)}^2}}& =a^{10}{b}^{10}\\ {\displaystyle \frac{b^5{c}^5}{a^2{b}^8}}& =a^{10}{b}^{10}\\ c^5& ={\displaystyle \frac{a^2{b}^8\cdot a^{10}{b}^{10}}{b^5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle a^{12}{b}^{13}}}}}\end{array}$$

Jadi, c⁵ = a¹²b¹³.
JAWAB: E

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No. 3

{\dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}=....}

1. −1
2. 0
3. \dfrac12

4. 1
5. A^{x + y}

Zenius.net

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{1}{1+A^{x-y}}}+{\displaystyle \frac{1}{1+A^{y-x}}}& ={\displaystyle \frac{1}{1+{\displaystyle \frac{A^x}{A^y}}}}+{\displaystyle \frac{1}{1+{\displaystyle \frac{A^y}{A^x}}}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{A^y+A^x}{A^y}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{A^x+A^y}{A^x}}}}\\ & ={\displaystyle \frac{A^y}{A^y+A^x}}+{\displaystyle \frac{A^x}{A^x+A^y}}\\ & ={\displaystyle \frac{A^y}{A^y+A^x}}+{\displaystyle \frac{A^x}{A^y+A^x}}\\ & ={\displaystyle \frac{A^y+A^x}{A^y+A^x}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$$

Jadi, \dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}=1.
JAWAB: D

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No. 4

Bentuk sederhana dari \left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2 adalah ....

1. \dfrac{p^6}{q^5r^4}
2. \dfrac{p^6}{q^6r^2}
3. \dfrac{p^3}{q^5r^2}

4. \dfrac{q^6}{p^6r^4}
5. \dfrac{q^5p^6}{r^2}

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\left({\displaystyle \frac{p^{-2}{q}^{2}r}{p{q}^{-1}{r}^3}}\right)}^2& ={\left({\displaystyle \frac{q^{2-(-1)}}{p^{1-(-2)}{r}^{3-1}}}\right)}^2\\ & ={\left({\displaystyle \frac{q^3}{p^3{r}^2}}\right)}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{q^6}{p^6{r}^4}}}}}}\end{array}$$

Jadi, \left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2=\dfrac{q^6}{p^6r^4}.
JAWAB: D

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No. 5

Hasil dari \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}} adalah

1. \dfrac{27}2
2. \dfrac92
3. \dfrac{27}8

4. \dfrac98
5. \dfrac8{27}

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{8^{-\frac{3}{5}}{9}^{\frac{5}{4}}}{{81}^{-\frac{1}{8}}{64}^{\frac{1}{5}}}}& ={\displaystyle \frac{8^{-\frac{3}{5}}{\left(3^2\right)}^{\frac{5}{4}}}{{\left(3^4\right)}^{-\frac{1}{8}}{\left(8^2\right)}^{\frac{1}{5}}}}\\ & ={\displaystyle \frac{8^{-\frac{3}{5}}{3}^{\frac{5}{2}}}{3^{-\frac{1}{2}}{8}^{\frac{2}{5}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{2}+\frac{1}{2}}}{8^{\frac{2}{5}+\frac{3}{5}}}}\\ & ={\displaystyle \frac{3^{\frac{6}{2}}}{8^{\frac{5}{5}}}}\\ & ={\displaystyle \frac{3^3}{8^1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{27}{8}}}}}}\end{array}$$

Jadi, \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}}=\dfrac{27}8.
JAWAB: C

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No. 6

2⁵ × 2⁷ = ....

1. 2⁸
2. 2¹²

3. 2¹⁸
4. 2²⁵

5. 2³⁵

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{2}^5\times 2^7& =2^{5+7}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{2}^{12}}}}}\end{array}$$

Jadi, 2⁵ × 2⁷ = 2¹².

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JAWAB: B

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No. 7

$\dfrac{3^5}{3^2}=$ ....

1. 1
2. 3

3. 9
4. 27

5. 81

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{3^5}{3^2}}& =3^{5-2}\\ & =3^3\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black27}}}}\end{array}$

Jadi, $\dfrac{3^5}{3^2}=27$.
JAWAB: D

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No. 8

Bentuk sederhana dari $\dfrac{7x^3y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}=$ ....

1. $\dfrac{x^{10}z^{10}}{12y^3}$
2. $\dfrac{z^2}{12x^4y^3}$

3. $\dfrac{x^{10}y^5}{12z^2}$
4. $\dfrac{y^3z^2}{12x^4}$

5. $\dfrac{x^{10}}{12y^3z^2}$

Big Book M3qatematika SMA

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{7x^3{y}^{-4}{z}^{-6}}{84x^{-7}{y}^{-1}{z}^{-4}}}& ={\displaystyle \frac{x^{3-(-7)}}{12y^{-1-(-4)}{z}^{-4-(-6)}}}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{\displaystyle \frac{x^{10}}{12y^3{z}^2}}}}}}\end{array}$$

Jadi, $\dfrac{7x^3y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}=\dfrac{x^{10}}{12y^3z^2}$.
JAWAB: E

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No. 9

Bentuk sederhana dari \dfrac{x^5\cdot y^7}{x^4\cdot y^8}

1. \dfrac{x}y
2. x²⋅ y

3. x⋅ y
4. x⋅ y²

Grup Telegram

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{x^5\cdot y^7}{x^4\cdot y^8}}& ={\displaystyle \frac{x^{5-4}}{y^{8-7}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{x}{y}}}}}}\end{array}$$

Jadi, \dfrac{x^5\cdot y^7}{x^4\cdot y^8}=\dfrac{x}y.
JAWAB: A

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No. 10

Sederhanakan bentuk dari \left(\dfrac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot2r^2}\right)^2

Grup WhatsApp

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\left({\displaystyle \frac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot 2r^2}}\right)}^2& ={\left({\displaystyle \frac{3q^{5-4}{r}^{3-2}}{p^{7-2}}}\right)}^2\\ & ={\left({\displaystyle \frac{3qr}{p^5}}\right)}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{6q^2{r}^2}{p^{10}}}}}}}\end{array}$$

Jadi, \left(\dfrac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot2r^2}\right)^2=\dfrac{6q^2r^2}{p^{10}}.

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