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`@` `\text{Ans}`
`\downarrow`
`2^(2x-4)=64`
`=>2^(2x-4)=2^6`
`=>2x-4=6`
`=>2x=10`
`=>x=10 \div 2`
`=> x=5`
Vậy, `x = 5.`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
`(3*x+2)^2=121`
\(=>\left[{}\begin{matrix}3x+2=11\\3x+2=-11\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=11-2\\3x=-11-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}3x=9\\3x=-13\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{13}{3}\left(loại\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3) Tìm số tự nhiên x, biết:
a) \(2^x=4\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
___________
b) \(2^x=1\)
\(2^x=2^0\)
\(\Rightarrow x=0\)
___________
c) \(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
___________
d) \(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
___________
e) \(5^x=125\)
\(5^x=5^3\)
\(\Rightarrow x=3\)
___________
f) \(8^x=64\)
\(8^x=8^2\)
\(\Rightarrow x=2\)
___________
h) \(3^{x+1}=3^2\)
\(\rightarrow x+1=2\)
\(x=2-1\)
\(x=1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
a: 2^x=4
=>2^x=2^2
=>x=2
b: 2^x=1
=>2^x=2^0
=>x=0
c: 2^x=16
=>2^x=2^4
=>x=4
d; 3^x=9
=>3^x=3^2
=>x=2
e: 5^x=125
=>5^x=5^3
=>x=3
f: 8^x=64
=>8^x=8^2
=>x=2
f: 3^x+1=3^2
=>x+1=2
=>x=1
`@` `\text {Ans}`
`\downarrow`
\(\left(2\cdot x+2\right)^2=64\)
`\Rightarrow`\(\left(2x+2\right)^2=\left(\pm8\right)^2\)
`\Rightarrow`\(\left[{}\begin{matrix}2x+2=8\\2x+2=-8\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}2x=8+2\\2x=-8+2\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}2x=10\\2x=-6\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=10\div2\\x=-6\div2\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy, `x \in {5; -3}`
`@` `\text {Kaizuu lv uuu}`