Exercise Zone : Pangkat (Eksponen) [2]

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Tipe:

Standar SBMPTN HOTS

No.

Sederhanakan dan tentukan nilai dari ${64}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{4}}\right)}^2\cdot {32}^{-\frac{3}{5}}$

Grup Telegram

$\begin{array}{rl}{64}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{4}}\right)}^2\cdot {32}^{-\frac{3}{5}}& ={\left(2^6\right)}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{2^2}}\right)}^2\cdot {\left(2^5\right)}^{-\frac{3}{5}}\\ & =2^2\cdot {\left(2^{-2}\right)}^2\cdot 2^{-3}\\ & =2^2\cdot 2^{-4}\cdot 2^{-3}\\ & =2^{2-4-3}\\ & =2^{-5}\\ & ={\displaystyle \frac{1}{2^5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{32}}}}}}\end{array}$

Jadi, ${64}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{4}}\right)}^2\cdot {32}^{-\frac{3}{5}}={\displaystyle \frac{1}{32}}$.

No.

Sederhanakanlah operasi perpangkatan di bawah ini

1. $5^5\times 5^{-4}$
2. $y^3\times {\left(3y\right)}^2$
3. ${\left({\displaystyle \frac{1}{2}}\right)}^3\times {\left({\displaystyle \frac{1}{4}}\right)}^2$

Grup Telegram

1. $5^5\times 5^{-4}=5^{5+(-4)}=5^1=5$
2. $y^3\times {\left(3y\right)}^2=y^3\times 3^2\times y^2=9y^5$
3. ${\left({\displaystyle \frac{1}{2}}\right)}^3\times {\left({\displaystyle \frac{1}{4}}\right)}^2={\displaystyle \frac{1}{8}}\times {\displaystyle \frac{1}{16}}={\displaystyle \frac{1}{128}}$

Jadi, $5$ $9y^5$ ${\displaystyle \frac{1}{128}}$

No.

Jika $K^{2x}=3$, maka ${\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}=$

1. ${\displaystyle \frac{36}{122}}$
2. ${\displaystyle \frac{39}{122}}$
3. ${\displaystyle \frac{81}{244}}$

4. ${\displaystyle \frac{122}{39}}$
5. ${\displaystyle \frac{244}{81}}$

$\begin{array}{rl}{\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}& ={\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}\cdot {\displaystyle \frac{K^{5x}}{K^{5x}}}\\ & ={\displaystyle \frac{K^{8x}-K^{2x}}{K^{10x}+K^0}}\\ & ={\displaystyle \frac{{\left(K^{2x}\right)}^4-K^{2x}}{{\left(K^{2x}\right)}^5+1}}\\ & ={\displaystyle \frac{3^4-3}{3^5+1}}\\ & ={\displaystyle \frac{81-3}{243+1}}\\ & ={\displaystyle \frac{78}{244}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{39}{122}}}}}}\end{array}$

Jadi, ${\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}=$. JAWAB: B

No.

Hasil dari $6^8:6^5$ adalah ....

SUMBER

$\begin{array}{rl}{6}^8:6^5& =6^{8-5}\\ & =6^3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 216}}}}\end{array}$

Jadi, $6^8:6^5=216$.

No.

Bentuk sederhana dari ${\displaystyle \frac{5a^2{b}^{-7}{c}^3}{10a^5{b}^3{c}^{-2}}}$ adalah ....

SUMBER

$\begin{array}{rl}{\displaystyle \frac{5a^2{b}^{-7}{c}^3}{10a^5{b}^3{c}^{-2}}}& ={\displaystyle \frac{c^{3-(-2)}}{2a^{5-2}{b}^{3-(-7)}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{c^5}{2a^3{b}^{10}}}}}}}\end{array}$

Jadi, bentuk sederhana dari ${\displaystyle \frac{5a^2{b}^{-7}{c}^3}{10a^5{b}^3{c}^{-2}}}$ adalah ${\displaystyle \frac{c^5}{2a^3{b}^{10}}}$.

No.

Bila $x=9$ dan $y=64$ maka nilai ${\displaystyle \frac{x^{-\frac{3}{2}}\sqrt[3]{y^2}}{y^{\frac{1}{3}}-x^{\frac{1}{2}}}}$ adalah ....

Latihan USBN SMK

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\frac{x^{-\frac{3}{2}}\sqrt[3]{y^2}}{y^{\frac{1}{3}}-x^{\frac{1}{2}}}& =\frac{9^{-\frac{3}{2}}\sqrt[3]{{64}^2}}{{64}^{\frac{1}{3}}-9^{\frac{1}{2}}}\\ & =\frac{{\left(3^2\right)}^{-\frac{3}{2}}{\left(4^3\right)}^{\frac{2}{3}}}{{\left(4^3\right)}^{\frac{1}{3}}-{\left(3^2\right)}^{\frac{1}{2}}}\\ & =\frac{3^{-3}{4}^2}{4-3}\\ & =\frac{\frac{1}{3^3}\cdot 16}{1}\\ & =\frac{1}{27}\cdot 16\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \frac{16}{27}}}}}\end{array}$$

Jadi, nilai ${\displaystyle \frac{x^{-\frac{3}{2}}\sqrt[3]{y^2}}{y^{\frac{1}{3}}-x^{\frac{1}{2}}}}$ adalah ${\displaystyle \frac{16}{27}}$.

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No.

Bentuk sederhana dari $(2x^2\cdot y^{-5})(-2x^{-8}\cdot y^9)$ adalah ....

Latihan USBN SMK

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}(2x^2\cdot y^{-5})(-2x^{-8}\cdot y^9)& =(2)(-2)x^{2+(-8)}\cdot y^{-5+9}\\ & =-4x^{-6}{y}^4\\ & =-\frac{4y^4}{x^6}\end{array}$$

Jadi, bentuk sederhana dari $(2x^2\cdot y^{-5})(-2x^{-8}\cdot y^9)$ adalah $-\frac{4y^4}{x^6}$.

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