Question

1) giai phuong trinh:

a) \(x+\sqrt{2x+3}=2x\left(x-2\right)\)

Answer 1

Lời giải:

ĐK: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow \sqrt{2x+3}=2x^2-5x$

$\Leftrightarrow \sqrt{2x+3}-3=2x^2-5x-3$

$\Leftrightarrow \frac{2(x-3)}{\sqrt{2x+3}+3}=(2x+1)(x-3)$

$\Leftrightarrow (x-3)\left[\frac{2}{\sqrt{2x+3}+3}-(2x+1)\right]=0$

Xảy ra 2 TH:

TH1: $x-3=0\Rightarrow x=3$ (thỏa mãn)

TH2: $\frac{2}{\sqrt{2x+3}+3}=2x+1$

Đặt $\sqrt{2x+3}=t(t\geq 0)$ thì pt trở thành: \frac{2}{t+3}=t^2-2$

$\Leftrightarrow 2=(t^2-2)(t+3)\Leftrightarrow t^3+3t^2-2t-8=0$

$\Leftrightarrow (t+2)(t^2+t-4)=0$

Do $t\geq 0$ nên $t=\frac{-1+\sqrt{17}}{2}$

$\Leftrightarrow \sqrt{2x+3}=\frac{-1+\sqrt{17}}{2}\Leftrightarrow x=\frac{3-\sqrt{17}}{4}$ (thỏa mãn)

Vậy........