$f(2x-5)=x^2+3x-2)$ Tìm $f(x)$

Ta có  f ( x ) = 0f (x) = 0
⇔ x 2−5 x + 4 = 0⇔x2−5x + 4 = 0
⇔ x 2−4 x – x + 4 = 0⇔x2−4x − x + 4 = 0
⇔ x ( x −4) – ( x −4) = 0⇔x (x − 4) – (x − 4) = 0
⇔ ( x −1) ( x −4) = 0⇔ (x− 1) (x − 4) = 0
⇔ x = 1⇔x = 1 hoặc  x = 4x ​​= 4
:. . .
b) f (x) = 2 x 2×2 + 3x + 1
Ta có  f ( x ) = 0f (x) = 0
⇔2 x 2 + 3 x + 1 = 0⇔2×2 + 3x + 1 = 0
⇔2x 2 + 2 x + x + 1 = 0⇔2×2 + 2x + x + 1 = 0
⇔2 x ( x +1) + ( x +1) = 0⇔2x (x + 1) + (x + 1) = 0
⇔ ( x +1) (2 x +1) = 0⇔ (x + 1) (2x + 1) = 0
⇔ x = −1⇔x = −1 hoặc  x = −12x = −12