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a) x = 4
b) x = 3
c) x = 2
d) x = 1
e) x = 3
f) x = 2
g) x = 4
h) x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2\left(x-51\right)=2\cdot2^3+20\)
=>\(2\left(x-51\right)=2^4+20=36\)
=>x-51=36/2=18
=>x=18+51=69
b: \(2x-49=5\cdot3^2\)
=>\(2x-49=5\cdot9=45\)
=>2x=45+49=94
=>x=94/2=47
c: \(\left[\left(8x-12\right):4\right]\cdot3^3=3^6\)
=>\(\left[4\cdot\dfrac{\left(2x-3\right)}{4}\right]=3^3\)
=>\(2x-3=3^3=27\)
=>2x=3+27=30
=>x=30/2=15
d: \(2^{x+1}-2^2=32\)
=>\(2^{x+1}=32+2^2=32+4=36\)
=>\(x+1=log_236\)
=>\(x=log_236-1\)
e: \(\left(x^3-77\right):4=5\)
=>\(x^3-77=20\)
=>\(x^3=77+20=97\)
=>\(x=\sqrt[3]{97}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Ta có: \(2^{x+1}\cdot3^y=12^x\)
\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(6^{2x-1}=216\)
\(\Leftrightarrow2x-1=3\)
hay x=2
c: Ta có: \(x^3=25x\)
\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,x^3=7^3\Rightarrow x=7\\ b,2^x=4=2^2\\ Vậy:x=2\)
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