[Toán 9] Phương trình khó

H

hien_vuthithanh

$$(x+1)^2(2x+1)(2x+3)=18$$
$$\iff (x^2+2x+1)(4x^2+8x+3)=18$$

Đặt $x^2+2x=t (t \ge -1)$

PT $$\iff (t+1)(4t+3)=18$$
$$\iff 4t^2+7t-15=0 \iff t= \begin{bmatrix}& t=\dfrac{5}{4} & \\ & t=-3 & \end{bmatrix} \Longrightarrow t=\dfrac{5}{4}$$

$$\iff x^2+2x=\dfrac{5}{4}$$

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