`@` `\text {Ans}`
`\downarrow`
`a)`
`(x - 2)(x - 3) + (x-2) - 1 = 0`
`=> x(x-3) - 2(x - 3) + x - 2 - 1 = 0`
`=> x^2 - 3x - 2x + 6 + x - 3 = 0`
`=> x^2 - (3x + 2x - x) + (6 - 3) = 0`
`=> x^2 - 4x + 3 = 0`
`=> x^2 - 3x - x + 3 = 0`
`=> (x^2 - x) - (3x - 3) = 0`
`=> x(x - 1) - 3(x - 1) = 0`
`=> (x-3)(x-1) = 0`
`=>`\(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 3}`
`b)`
`(x + 2)^2 - 2x(2x + 3) = (x+1)^2`
`=> (x^2 + 4x + 4) - 4x^2 - 6x = x^2 + 2x + 1`
`=> x^2 + 4x + 4 - 4x^2 - 6x = x^2 + 2x + 1`
`=> -3x^2 - 2x + 4 = x^2 + 2x + 1`
`=> -3x^2 - 2x + 4 - x^2 - 2x - 1 = 0`
`=> -4x^2 - 4x + 3 = 0`
`=> -4x^2 - 6x + 2x + 3 = 0`
`=> -(4x^2 + 6x) + (2x + 3) = 0`
`=> -2x(2x + 3) + (2x + 3) = 0`
`=> (-2x + 1)(2x + 3) = 0`
`=>`\(\left[{}\begin{matrix}-2x+1=0\\2x+3=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}-2x=-1\\2x=-3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{-1}{-2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `x \in {-3/2; 1/2}.`
