(3x-1)3=-8/27
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\(\left(3x-1\right)^3=-\dfrac{8}{27}\\ \Leftrightarrow3x-1=-\dfrac{2}{3}\\ \Leftrightarrow3x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow3x-1=-\dfrac{2}{3}\Rightarrow3x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(3^{x+1}=27\)
\(\Leftrightarrow3^{x+1}=3^3\)
\(\Leftrightarrow x+1=3\Leftrightarrow x=2\)
2) \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=3\Leftrightarrow2x=4\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)
\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x/-15=-60/x
x2012=2012
(3x-1)3=-8/27
-2/x=-x/8/25
(8x-1)2n=52n
(x-2/3)3=(2/3)6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-1\right)^3=\frac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{2}{3}\)
\(\Rightarrow3x=\frac{5}{3}\)
\(\Rightarrow x=\frac{5}{3}:3\)
\(\Rightarrow x=\frac{5}{9}\)
tíc mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a/\left(3x+\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3\Leftrightarrow3x+\frac{1}{2}=-\frac{3}{2}\Leftrightarrow x=-\frac{2}{3}\)
\(b/-3x^2+15=0\Leftrightarrow3\left(5-x^2\right)=0\Leftrightarrow5-x^2=0\Leftrightarrow x=\pm\sqrt{5}\)
ta có : \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-1=-\frac{2}{3}\)
=> \(3x=\frac{1}{3}\)
=> \(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)