Chủ đề này có 2 trả lời

[danganhaaaa]

GPT

$10x^{2}-x$+6=2(2x+1)$\sqrt{2x^{2}-x+4}$

[triethuynhmath]

GPT

$10x^{2}-x$+6=2(2x+1)$\sqrt{2x^{2}-x+4}$

Chém bài này:
DKXD: $2x^2-x+4\geq 0$
PT $\Leftrightarrow 2x^2-x+4-2(2x+1)\sqrt{2x^2-x+4}+2(4x^2+1)=0\Leftrightarrow (\sqrt{2x^2-x+4}-2x-1)^2+(2x-1)^2=0\Leftrightarrow \left\{\begin{matrix} x=\frac{1}{2} \\ \sqrt{2x^2-x+4}=2x+1 \end{matrix}\right.\Leftrightarrow x=\frac{1}{2}(Q.E.D)$

[nthoangcute]

$10x^{2}-x+6=2(2x+1)\sqrt{2x^{2}-x+4}$

$$10x^{2}-x+6=2(2x+1)\sqrt{2x^{2}-x+4}$$
$$\Leftrightarrow \left\{\begin{matrix}
(10x^2-x+6)^2=4(2x+1)^2(2x^2-x+4)\\
2x+1 \geq 0
\end{matrix}\right.$$

$$\Leftrightarrow \left\{\begin{matrix}
(17x^2+8x+20)(2x-1)^2=0\\
2x+1 \geq 0
\end{matrix}\right.$$
$$\Leftrightarrow x=\frac{1}{2}$$