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Tìm x : (52x . 5x+1) :5 = 125
=> 5x ( 52 . 51 ) = 625
=> 5x . 125 = 625
=> 5x = 5
=> x = 1
Study well
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`@` `\text {Ans}`
`\downarrow`
`a)`
\(5\cdot x^3-5=0\)
`=> 5*x^3 = 0+5`
`=> 5*x^3 = 5`
`=> x^3 = 5 \div 5`
`=> x^3 = 1`
`=> x^3 = 1^3`
`=> x=1`
Vậy, `x=1.`
`b)`
\(( x+1)^2 = 16\)
`=> (x+1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy, `x \in {3; -5}`
`c)`
\(( x+1)^3 = 27\)
`=> (x+1)^3 = 3^3`
`=> x+1=3`
`=> x=3-1`
`=> x=2`
Vậy, `x=2.`
`d)`
\(( x-1)^3 = 343\)
`=> (x-1)^3 = 7^3`
`=> x-1=7`
`=> x=7+1`
`=> x=8`
Vậy, `x=8.`
`e)`
\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?
Mình làm cả 2 TH nhé!
`(2x-1^3)=125`
`=> 2x-1=125`
`=> 2x=125+1`
`=> 2x=126`
`=> x=126 \div 2`
`=> x=63`
TH2:
`(2x-1)^3 = 125`
`=> (2x-1)^3 = 5^3`
`=> 2x-1=5`
`=> 2x=5+1`
`=> 2x=6`
`=> x=6 \div 2`
`=> x=3`
Vậy, `x=3.`
(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)
(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)
(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)
(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 8 + (x - 9) = 125 - 64
8 + (x - 9) = 61
x - 9 = 53
x = 62
b. 5 x (X + 7) - 10 = 8 x 5
5 x (X + 7) - 10 = 40
5 x (X + 7) = 50
X + 7 = 10
X = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(\left(x-6\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=9\end{matrix}\right.\)
b,
\(\left|x\right|=3\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
c,
\(\left|x+5\right|=15\\ \Rightarrow\left[{}\begin{matrix}x+5=-15\\x+5=15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-20\\x=10\end{matrix}\right.\)
d,
\(2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
e,
\(5^{x+1}=125\\ \Rightarrow5^{x+1}=5^3\\ \Rightarrow x+1=3\\ \Rightarrow x=2\)
a: Ta có: \(\left(x-6\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)
b: Ta có: \(\left|x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: Ta có: \(\left|x+5\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-20\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,x=0 hoặc 1
b,x=5
c,x=8
d,x=2
e,x=9
g,x=2
f,bạn ghi rõ đề ra chứ mình không hiểu 1 số số là cơ số hay số mũ
(5^2.x-5^x+1)/5=125
25.x-5^x+1=625