a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

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.

a) \(3\left(2x-5\right)+125=134\)

\(\Leftrightarrow3\left(2x-5\right)=9\)

\(\Leftrightarrow2x-5=3\)

\(\Leftrightarrow2x=8\Leftrightarrow x=4\)

b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)

\(\Leftrightarrow6x+9=27\)

\(\Leftrightarrow6x=18\Leftrightarrow x=3\)

d) \(27\left(x-27\right)-27=0\)

\(\Leftrightarrow27\left(x-27\right)=27\)

\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)

3.(x+5)-2x=27
3.x+3.5-2x=27
3x+15-2x=27
x.(3-2)+15=27
x.  1    +15=27
  x       +15=27
  x             =27-15
  x             =12

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

\(\Rightarrow1-2x=-2\)

\(\Rightarrow2x=3\)

\(\Rightarrow=\frac{3}{2}\)

\(b,\left(2x-1\right)^3=-27\)

\(\Rightarrow2x-1=-3\)

\(\Rightarrow2x=-2\)

\(\Rightarrow x=-1\)

a/\(x:27=3,6\)
\(\Rightarrow x=97,2\)
b/\(\dfrac{2x+1}{-27}=\dfrac{-3}{2x+1}\)
\(\Rightarrow\left(2x+1\right)^2=81\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)

a) x=3,6\(\times\) 27=97,2

a,x:27=3,6

x=3,6:27

x\(\approx\)0,13

\(\Rightarrow\)x=0,1(3)

a. (x - 2)² = 1

<=> (x - 2)² = 1² = (-1)²

<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)

Vậy x \(\in\){1; 3}.

b. (2x - 1)³ = -8

<=> (2x - 1)³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -2 + 1

<=> 2x = -1

<=> x = -1/2

Vậy x = -1/2.

c. (x + 1/2)² = 1/16

<=> (x + 1/2)² = (1/4)² = (-1/4)²

<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)

Vậy x \(\in\){-1/4; -3/4}.

d. (x - 2)³ = -27

<=> (x - 2)³ = (-3)³

<=> x - 2 = -3

<=> x = -3 + 2

<=> x = -1

Vậy x = -1.

a.\(\left(x-2\right)^2\)=1

<=> x-2=1 hoặc x-2=-1

<=> x= 3 hoặc x=1

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

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

2x-1=-2

2x=-1

x=-1/2

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

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

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

x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)

d.\(\left(x-2\right)^3\)=-27

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

x-2=-3

x=-1

g.\(3^{x-1}\)=\(\frac{1}{243}\)

\(3^{x-1}\)=\(\frac{1}{3^5}\)

\(3^{x-1}\)=\(3^{-5}\)

x-1=-5

x=-4