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

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

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

\(x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy x = 0 hoặc x = -1

\(\left(2x+1\right)^2=25\)

\(\Leftrightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)

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

Vậy x = 2 hoặc x = -3

\(\left(2x-3\right)^2=36\)

\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4,5\\x=-1,5\end{cases}}\)

Vậy x = 4,5 hoặc x = -1,5

\((2x+1)^2=25\)

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

\(\Rightarrow(2x+1)=\pm5\)

Tự xét :v

\((2x-3)^2=36\)

\(\Rightarrow(2x-3)^2=6^2\)

\(\Rightarrow(2x-3)=\pm6\)

\(\Rightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=-\frac{3}{2}\end{cases}}\)

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

2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)

3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)

4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)

1) ( 2x + 1 )² = 25

=> ( 2x + 1 )² = 5²

=> 2x + 1 = 5 hoặc 2x + 1 = -5

=> 2x = 4 hoặc 2x = -6

=> x = 2 hoặc x = -3

2) 5^x+2 = 625

=> 5^x+2 = 5⁴

=> x + 2 = 4

=> x = 2

3) ( 2x - 3 )² = 36

=> ( 2x - 3 )² = 6²

=> 2x - 3 = 6 hoặc 2x - 3 = -6

=> 2x = 9 hoặc 2x = -3

=> x = 9/2 hoặc x = -3/2

4) ( 2x - 1 )³ = -8

=> ( 2x - 1 )³ = ( -2 )³

=> 2x - 1 = -2

=> 2x = -1

=> x = -1/2

Bài 1: 

a) Ta có: \(-5x+32=\left(-2\right)^3\)

\(\Leftrightarrow-5x+32=-8\)

\(\Leftrightarrow-5x=-40\)

hay x=8

Vậy: x=8

b) Ta có: \(-2x+36=6\)

\(\Leftrightarrow-2x=6-36=-30\)

hay x=15

Vậy: x=15

e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

b,-2x+36=6

tương đương -2x=-30

tương đương x=15

a, -5x+32=(-2)^3

tương đương -5x+32=8

tương đương -5x=-24

tương đương x=24/5

Bài làm

a) x² - 3 = 22

=> x² = 25

=> x = + 5

Vậy x = + 5

b) 2x³ + 5 = -11

2x³ = -16

x³ = -8

x = -2

Vậy x = -2

c) ( x + 2 )² = 81

=> x + 2 = 9

=> x = 7

Vậy x = 7

d) ( 2x + 1 )² = 25

=> 2x + 1 = 5

=> 2x = 4

=> x = 2

Vậy x = 2

e) 5^x + 2 = 625

5^x = 623 ( vô lí )

g) ( 2x - 3 )² = 36.

=> 2x - 3 = 6

=> 2x = 9

=> x = 4,5

Vậy x = 4,5

h) ( 2x - 1 )³ = -8

=> 2x - 1 = -2

=> 2x = -1

=> x = -1/2

Vậy x = -1/2

i) ( x - 1 )^x + 2 =  ( x - 1 )^x + 6

=> [ (x - 1 )^x - ( x - 1 )^x ] = 6 - 2

=> 0 = 4 ( vô lí )

Vậy x thuộc rỗng.

k) x² + x = 0

=> x( x + 1 ) = 0

=> x = 0 hoặc x + 1 = 0

=> x = 0 hoặc x = -1

Vậy x = 0 hoặc x = -1