(2X+1)^3=125
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Ta có \(\left(-\frac{1}{3}+2x\right)^3=\frac{-1}{125}\)
\(\Rightarrow\left(-\frac{1}{3}+2x\right)^3=-\left(\frac{1}{5}\right)^3\)
\(\Rightarrow2x=-\frac{1}{5}+\frac{1}{3}\)
\(\Rightarrow2x=\frac{2}{15}\)
\(\Rightarrow x=\frac{1}{15}\)
\(\Rightarrow2x=-\frac{1}{5}+\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: ( 2 x + 1 ) 3 = 3 3 nên 2x + 1 = 3. Do đó x = 1.
b) Ta có: ( 2 x - 1 ) 3 = 5 3 nên 2x - 1 = 5. Do đó x = 3.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow2x+1=5\)
hay x=2
Giải:
\(\left(2x+1\right)^3=125\)
\(\Rightarrow\) \(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow4:2\)
Vậy \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^3=125\)
\(\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-1\right)^3=125\\ \Rightarrow2x-1=5\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x+1)3 = 125
<=> (2x+1)3 = 53
<=> 2x+1 = 5
<=> 2x = 4
<=> x = 2
\(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=\dfrac{4}{2}\)
\(\Rightarrow x=2\)
Vậy: x=2
( 2x + 1 ) mũ 3 = 125
( 2x + 1 ) mũ 3 = 5 mũ 3
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2