Exercise Zone : Pangkat (Eksponen) [2]
Table of Contents
Berikut ini adalah kumpulan soal mengenai Pangkat. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.
Tipe:
No.
Sederhanakan dan tentukan nilai dari \({64^{\frac26}\cdot\left(\dfrac14\right)^2\cdot32^{-\frac35}}\)ALTERNATIF PENYELESAIAN
\begin{aligned}
64^{\frac26}\cdot\left(\dfrac14\right)^2\cdot32^{-\frac35}&=\left(2^6\right)^{\frac26}\cdot\left(\dfrac1{2^2}\right)^2\cdot\left(2^5\right)^{-\frac35}\\
&=2^2\cdot\left(2^{-2}\right)^2\cdot2^{-3}\\
&=2^2\cdot2^{-4}\cdot2^{-3}\\
&=2^{2-4-3}\\
&=2^{-5}\\
&=\dfrac1{2^5}\\
&=\boxed{\boxed{\dfrac1{32}}}
\end{aligned}
Jadi, \({64^{\frac26}\cdot\left(\dfrac14\right)^2\cdot32^{-\frac35}=\dfrac1{32}}\).
No.
Sederhanakanlah operasi perpangkatan di bawah ini55 × 5−4
ALTERNATIF PENYELESAIAN
55 × 5−4 = 55 + (−4) = 51 = 5
Jadi, 55 × 5−4 = 5.
No.
Jika- \(\dfrac{36}{122}\)
- \(\dfrac{39}{122}\)
- \(\dfrac{81}{244}\)
- \(\dfrac{122}{39}\)
- \(\dfrac{244}{81}\)
ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}&=\dfrac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}{\color{red}\cdot\dfrac{K^{5x}}{K^{5x}}}\\
&=\dfrac{K^{8x}-K^{2x}}{K^{10x}+K^0}\\
&=\dfrac{\left(K^{2x}\right)^4-K^{2x}}{\left(K^{2x}\right)^5+1}\\
&=\dfrac{3^4-3}{3^5+1}\\
&=\dfrac{81-3}{243+1}\\
&=\dfrac{78}{244}\\
&=\boxed{\boxed{\dfrac{39}{122}}}
\end{aligned}
Jadi, \(\dfrac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}=\dfrac{39}{122}\).
JAWAB: B
JAWAB: B
No.
Hasil dariALTERNATIF PENYELESAIAN
\begin{aligned}
6^8:6^5&=6^{8-5}\\
&=6^3\\
&=\boxed{\boxed{216}}
\end{aligned}
Jadi, 68 : 65 = 216 .
No.
Bentuk sederhana dari \(\dfrac{5a^2b^{-7}c^3}{10a^5b^3c^{-2}}\) adalah ....ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{5a^2b^{-7}c^3}{10a^5b^3c^{-2}}&=\dfrac{c^{3-(-2)}}{2a^{5-2}b^{3-(-7)}}\\
&=\boxed{\boxed{\dfrac{c^5}{2a^3b^{10}}}}
\end{aligned}
Jadi, bentuk sederhana dari \(\dfrac{5a^2b^{-7}c^3}{10a^5b^3c^{-2}}\) adalah \(\dfrac{c^5}{2a^3b^{10}}\).
No.
BilaALTERNATIF PENYELESAIAN
\begin{aligned}
\frac{x^{-\frac32}\sqrt[3]{y^2}}{y^{\frac13}-x^{\frac12}}&=\frac{9^{-\frac32}\sqrt[3]{64^2}}{64^{\frac13}-9^{\frac12}}\\[3pt]
&=\frac{\left(3^2\right)^{-\frac32}\left(4^3\right)^{\frac{2}{3}}}{\left(4^3\right)^{\frac13}-\left(3^2\right)^{\frac12}}\\[3pt]
&=\frac{3^{-3}4^2}{4-3}\\[3pt]
&=\frac{\frac1{3^3}\cdot16}1\\[3pt]
&=\frac1{27}\cdot16\\
&=\boxed{\boxed{\frac{16}{27}}}
\end{aligned}
Jadi, nilai $\dfrac{x^{-\frac32}\sqrt[3]{y^2}}{y^{\frac13}-x^{\frac12}}$ adalah \(\dfrac{16}{27}\).
No.
Bentuk sederhana dariALTERNATIF PENYELESAIAN
\begin{aligned}
\left(2x^2\cdot y^{-5}\right)\left(-2x^{-8}\cdot y^9\right)&=(2)(-2)x^{2+(-8)}\cdot y^{-5+9}\\
&=-4x^{-6}y^{4}\\
&=-\frac{4y^4}{x^6}
\end{aligned}
Jadi, bentuk sederhana dari (2x2 ⋅ y−5)(−2x−8 ⋅ y9) adalah \(-\frac{4y^4}{x^6}\).
No.
JikaALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{a^2b^3}{c^4}&=\frac{2^2\cdot3^3}{4^4}\\[3.7pt]
&=\frac{4\cdot27}{256}\\
&=\boxed{\boxed{\frac{27}{64}}}
\end{aligned}
Jadi, \(\dfrac{a^2b^3}{c^4}=\dfrac{27}{64}\).
No.
Hasil dariALTERNATIF PENYELESAIAN
\begin{aligned}
\left(3^2\right)^2\cdot\left(3^3\right)^{-3}&=3^4\cdot3^{-9}\\
&=3^{4+(-9)}\\
&=3^{-5}\\
&=\dfrac1{3^5}\\[3.6pt]
&=\dfrac1{243}
\end{aligned}
Jadi, hasil dari (32)2 ⋅ (33)−3 adalah \(\dfrac1{243}\).
No.
Jika a = 4, b = 3, dan c = 2, nilai dari \(\dfrac{a^2\cdot b^3}{c^4}\) adalah ....- 3
- 9
- 18
- 27
- 36
ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{a^2\cdot b^3}{c^4}&=\dfrac{4^2\cdot 3^3}{2^4}\\[3.6pt]
&=\dfrac{16\cdot 27}{16}\\
&=\boxed{\boxed{27}}
\end{aligned}
Jadi, nilai dari \(\dfrac{a^2\cdot b^3}{c^4}\) adalah 27.
JAWAB: D
JAWAB: D
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