\(\left(2x-1\right)^3=125\)

\(\Leftrightarrow\left(2x-1\right)^3=5^3\)

\(\Leftrightarrow2x-1=5\Leftrightarrow x=3\)

             \(------\)

\(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)

a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

Ta có \(\left(-\frac{1}{3}+2x\right)^3=\frac{-1}{125}\)

\(\Rightarrow\left(-\frac{1}{3}+2x\right)^3=-\left(\frac{1}{5}\right)^3\)

\(\Rightarrow2x=-\frac{1}{5}+\frac{1}{3}\)

\(\Rightarrow2x=\frac{2}{15}\)

\(\Rightarrow x=\frac{1}{15}\)

\(\Rightarrow2x=-\frac{1}{5}+\frac{1}{3}\)

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.

\(\left(2x+1\right)^3=125\)

\(\Leftrightarrow2x+1=5\)

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 \(x=2\)

\(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=\dfrac{4}{2}\)

\(\Rightarrow x=2\)

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

\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)

???????

\(\left(2x-1\right)^4=16\)

\(\left(2x-1\right)^4=2^4\)

\(2x-1=2\)

\(2x=3\)

\(x=\dfrac{3}{2}\)

\(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

\(\left(2x-1\right)^3=125\\ \Rightarrow2x-1=5\\ \Rightarrow2x=6\\ \Rightarrow x=3\)

(2x−1)³ = 125

⇒ 2x − 1 = 5

⇒ 2x = 6

⇒ x = 3

(2x−1)\(^3\) = 125 

(2x - 1)\(^3\) = 5\(^3\)

=> 2x - 1 = 5

      2x      = 5 + 1

       2x      = 6

         x      = 6 : 2

          x      = 3 

Vây x = 3

(2x+1)3 = 125 

<=> (2x+1)3 = 53

<=> 2x+1 = 5

<=> 2x = 4

<=> x = 2