Chủ đề này có 2 trả lời

[xuantrandong]

1.$\left ( x+4 \right )^{2}-6\sqrt{x^{3}+3x}=13$
2.${\sqrt{x-1}}+\sqrt{3-x}=3x^{2}-4x-2$
3.$\sqrt{x+2}+x^{2}-x-2\leq \sqrt{3x-2}$
4.$2\sqrt[3]{3x-2}+3\sqrt{6-5x}-8=0$
5.$\frac{x-\sqrt{x}}{1-\sqrt{2(x^{2}-x+1)}}\geq 1$
6.$\sqrt{3x+1}-\sqrt{6-x}+3x^{2}-14x-8=0$
7.$3\sqrt{2+x}-6\sqrt{2-x}+4\sqrt{4-x^{2}}=10-3x$
8.$\frac{8-x^{2}}{\sqrt{1+x}+\sqrt{1-x}}=4$

[dorabesu]

2.${\sqrt{x-1}}+\sqrt{3-x}=3x^{2}-4x-2$

http://diendantoanhoc.net/index.php?/topic/90612-gi%E1%BA%A3i-ph%C6%B0%C6%A1ng-trinh-sqrt3-xsqrtx-13x2-4x-2/
P/s : Lần sau bạn post ít thôi nhé, nhìn nhiều quá sợ lắm ^^

[Nguyen Duc Thuan]

6.$\sqrt{3x+1}-\sqrt{6-x}+3x^{2}-14x-8=0$

$\Leftrightarrow \sqrt{3x+1}-4-\left ( \sqrt{6-x}-1 \right )+3x^2-14x-5=0$
$\Leftrightarrow \frac{3(x-5)}{4+\sqrt{3x+1}}+\frac{x-5}{1+\sqrt{6-x}}+(x-5)(3x+1)=0$
$\Leftrightarrow (x-5)M=0 (M>0)$
$\Leftrightarrow x=5\square$