\(\left(3x-1\right)^3=-\dfrac{8}{27}\\ \Leftrightarrow3x-1=-\dfrac{2}{3}\\ \Leftrightarrow3x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)

\(\Leftrightarrow3x-1=-\dfrac{2}{3}\)

hay x=1/9

\(\left(3x-1\right)^3=-\left(\dfrac{2}{3}\right)^3\\ 3x-1=-\dfrac{2}{3}\\ 3x=-\dfrac{2}{3}+1\\ 3x=\dfrac{5}{3}\\ x=\dfrac{5}{3}:3\\ x=\dfrac{5}{9}\)

1) \(3^{x+1}=27\)

\(\Leftrightarrow3^{x+1}=3^3\)

\(\Leftrightarrow x+1=3\Leftrightarrow x=2\)

2) \(\left(2x-1\right)^3=8\)

\(\Leftrightarrow\left(2x-1\right)^3=2^3\)

\(\Leftrightarrow2x-1=3\Leftrightarrow2x=4\Leftrightarrow x=2\)

3^x+1=27
3^x+1=3³
x+1=3
x=2
(2x-1)³=8
(2x-1)³=2³
2x-1=2
2x=3
x=3/2

\(3^{x+1}=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

\(\left(2x-1\right)^3=8\)

\(\Leftrightarrow2x-1=2\)

hay \(x=\dfrac{3}{2}\)

\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)

\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)

\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)

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 \(x=\frac{1}{9}\)

x/-15=-60/x

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

\(\left(3x-1\right)^3=-\frac{8}{27}\)

\(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)

\(3x-1=-\frac{2}{3}\)

\(3x=-\frac{2}{3}+1\)

\(3x=-\frac{2}{3}+\frac{3}{3}\)

\(3x=-\frac{1}{3}\)

\(x=-\frac{1}{3}\div3\)

\(x=-\frac{1}{3}\times\frac{1}{3}\)

\(x=-\frac{1}{9}\)

\(\left(3x-1\right)^3=\frac{8}{27}\)

\(\Rightarrow\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)

\(\Rightarrow3x-1=\frac{2}{3}\)

\(\Rightarrow3x=\frac{5}{3}\)

\(\Rightarrow x=\frac{5}{3}:3\)

\(\Rightarrow x=\frac{5}{9}\)

tíc mình nha

3x=8/27+1

3x=35/27

x=35/27:3

x=35/81

x = 5 / 9

\(a/\left(3x+\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3\Leftrightarrow3x+\frac{1}{2}=-\frac{3}{2}\Leftrightarrow x=-\frac{2}{3}\)

\(b/-3x^2+15=0\Leftrightarrow3\left(5-x^2\right)=0\Leftrightarrow5-x^2=0\Leftrightarrow x=\pm\sqrt{5}\)