[2x +1] ³ =125

[2x +1] ³ =5³

2x +1         =5

2x               =4

x                 =4:2

x                 =2

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

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

\(2x+1=5\)

\(2x=5-1\)

2x=4

x=4/2

x=2

\(\left(4x-1\right)^2=25\cdot9\)

\(\left(4x-1\right)^2=225\)

\(\left(4x-1\right)^2=15^2\)

\(4x-1=15\)

4x=15+1

4x=16

x=16/4

x=4

(2x + 1)³ = 125

(2x + 1)³ = 5³

=> 2x + 1 = 5

     2x       = 4

       x       = 2

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

=>2x+1=5

2x=4

x=2

hok tốt

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

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

\(\Rightarrow2x+1=\pm5\)

\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy \(x\in\left\{2;-3\right\}\)

a) -3/5

b) -9/4

c) x thuộc N*( chắc thế)

Bn giải kĩ đc k 

Bài 3:

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(\Rightarrow x-2=\pm1\)

+) \(x-2=1\Rightarrow x=3\)

+) \(x-2=-1\Rightarrow x=1\)

Vậy \(x=3\) hoặc \(x=1\)

c) \(\left(2x-1\right)^3=-8\)

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

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

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

Vạy \(x=\frac{-1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

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

\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)

\(\Rightarrow x=\frac{-1}{4}\)

Vậy \(x=\frac{-1}{4}\)

suy ra 1/2+2x=0(1)hay2x-3=0(2)
giải(1)1/2+2x=0                   giải(2)2x-3=0
2x=0-1/2                               2x=0+3
2x=-1/2                                 2x=3
x=-1/2:2                                x=3:2
x=-1/4                                   x=3/2
    vẫy  x  ϵ {-1/4;3/2}

Sẽ có 2 trường hợp xảy ra

Trường hợp 1:

\(\dfrac{1}{2}\) + 2x = 0

       2x = 0 - \(\dfrac{1}{2}\)

       2x = -\(\dfrac{1}{2}\)

         x = -0,25

Trường hợp 2: 

2x - 3 = 0

2x      = 0 + 3

2x      = 3

  x      = 3:2

  x      = 1,5

=> x = -0,25 hoặc 1,5

\(x+\dfrac{1}{x}=3\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\\ \Leftrightarrow x^3+\left(\dfrac{1}{x}\right)^3+3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}+3\cdot3=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}=18\)

a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.