b)   ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0

<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0

<=> ( 2x - 3 )( 1 + x - 1 ) = 0

<=> x( 2x - 3 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

Vậy .....

a, 25x^2 - 1 - (5x -1)(x+2)=0

=> (5x)^2 - 1 + (5x-1)(x+2) = 0

=> (5x-1)(5x+1) + (5x-1)(x+2) = 0

=> (5x-1)(5x+1+x+2) = 0

=> (5x-1)(6x+3) = 0

=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)

a)     25x² - 1 - ( 5x - 1 )( x + 2 ) = 0

<=> ( 5x - 1 )( 5x + 1 ) - ( 5x - 1 )( x + 2 ) = 0

<=> ( 5x - 1 )( 5x + 1 - x - 2) = 0

<=> ( 5x - 1 )( 4x - 1 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{4}\end{cases}}}\)

Vậy .......

(4x-1)²-(x²+2x+1)=0

(4x-1)²-(x+1)²=0

(4x-1-x-1)(4x-1+x+1)=0

(3x-2)5x=0

x=0 hoặc 3x-2=0=>x=2/3

\(\left(4x-1\right)^2-x^2-2x-1=0\)

\(\Rightarrow\left(4x-1\right)^2-\left(x^2+2x+1\right)=0\)

\(\Rightarrow\left(4x-1\right)^2-\left(x+1\right)^2=0\)

\(\Rightarrow\left(4x-1-x-1\right)\left(4x-1+x+1\right)=0\)

\(\Rightarrow\left(3x-2\right).5x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) \(4x^2-1-x\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-x\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

b) \(\left(4x-1\right)^2-9=0\)

\(\Leftrightarrow\left(4x-1-3\right)\left(4x-1+3\right)=0\)

\(\Leftrightarrow\left(4x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}\)

\(\)

a, x=-1/2 hoặc 1

b, x=-1/2 hoặc x=1

P/s sai thì sửa nha

b)=(4x-1)²-3²=(4x-1-3)(4x-1+3)=16x^2-8x-8

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b: 

1: \(\Leftrightarrow2x\left(x+2\right)=0\)

=>x=0 hoặc x=-2

\((4x-1)^2-4(2x-3)^2-x-4=0\\\Leftrightarrow (4x)^2-2\cdot4x\cdot1+1^2-4[(2x)^2-2\cdot2x\cdot3+3^2]-x-4=0\\\Leftrightarrow 16x^2-8x+1-4(4x^2-12x+9)-x-4=0\\\Leftrightarrow 16x^2-8x+1-16x^2+48x-36-x-4=0\\\Leftrightarrow (16x^2-16x^2)+(-8x+48x-x)+(1-36-4)=0\\\Leftrightarrow 39x-39=0\\\Leftrightarrow 39x=39\\\Leftrightarrow x=1\\Vậy:x=1\)

(4x - 1)² -4(2x - 3)² - x - 4 = 0

16x² - 8x + 1 - 4(4x² - 12x + 9) - x - 4 = 0

16x² - 8x + 1 - 16x² + 48x - 36 - x - 4 = 0

39x - 39 = 0

39x = 39

x = 39 : 39

x = 1

<=> 8x^2 + 6x - 4x-3 + 4x-8x^2+8-16x=0

<=> 8x^2-8x^2+6x-4x+4x-16x = 3-8

<=> -10x= -5

<=> x=0,5

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1