Giải các phương trình: 
$1.\,\left(3x+1\right)\sqrt{2x^2-1}=5x^2+\dfrac{3}{2}x-3\\2.\,6x^2-10x+5-\left(4x-1\right)\sqrt{6x^2-6x+5}=0\\3.\,x^2+2x-2\sqrt{x+2}=4$ 
uh, t cũng thấy thế –  cuungonghinh 02-11-12 09:35 PM
trước tiên cứ tim đk cái đã –  conheodat297 31-10-12 09:32 PM
3. Điều kiện $x \ge -2$
$x^2+2x-2\sqrt{x+2}=4\Leftrightarrow x^2+2x-8-2(\sqrt{x+2}-2)=0$ 
$\Leftrightarrow (x+4)(x-2)-2(x-2)\frac{2}{\sqrt{x+2}+2}=0$ 
Do $x+4 \ge 2 >1\ge \frac{2}{\sqrt{x+2}+2}$.
Do vậy PT $\Leftrightarrow x=2.$ 
bài này giải chi tiết hơn nữa đc ko b? –  giacmotrua297 31-10-12 08:48 PM
2.
Điều kiện: $6x^2-6x+5\ge0$
Đặt: $\sqrt{6x^2-6x+5}=t,t\ge0$.
Phương trình trở thành:
     $t^2-(4x-1)t-4x=0$
$\Leftrightarrow (t-4x)(t+1)=0$
$\Leftrightarrow t=4x$
$\Leftrightarrow \sqrt{6x^2-6x+5}=4x$
$\Leftrightarrow \left\{ \begin{array}{l} x\ge0\\6x^2-6x+5=16x^2 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x\ge0\\10x^2+6x-5=0 \end{array} \right.\Leftrightarrow x=\frac{1}{10}(\sqrt{59}-3)$

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