1/ $\sqrt{x^{2}-\frac{7}{x^{2}}} +\sqrt{x-\frac{7}{x^{2}}}=0$
2/ $\sqrt{x-\sqrt{x^{2}-1}}+\sqrt{x+\sqrt{x^{2}-1}}=\sqrt{2(x^{3}+1)}$
3/ $\sqrt{x-5}= \frac{36}{\sqrt{x-5}}-\sqrt{x-4}$
4/ $x+\sqrt{x^{2}+16}=\frac{40}{\sqrt{x^{2}+16}}$
5/ $\sqrt{11x+3}-\sqrt{2-x}=\sqrt{9x+7}-\sqrt{x-2}$
Bạn chú ý trong cách nhập công thức nhé. Chỉ cần nhập 2 dấu $ một lần thôi, sau đó nhập công thức vào giữa hai dấu đó thì công thức mới hiển thị đúng dc. (BQT) thank! –  Trúc Anh 24-05-13 09:21 AM
4.$pt\Leftrightarrow x=\frac{24-x^2}{\sqrt{x^2+16}}=>x\sqrt{x^2+16}=24-x^2$
binh phuong hai ve 
$=>x^4-48x^2+576=x^4+16x^2=>x=^+_-3$
thay lai $=>x=3$
5.dk $x=2$.thay x=2 thay tm kl...
3.dk $x>5$ quy dong khu mau ta co 
$\sqrt{(x-5)(x-4)}=41-x\Leftrightarrow \begin{cases}x\leq 41\\ x^2-9x+20=x^2-82x+ 1681\end{cases}$
$=>$$x=\frac{1661}{73}$
 
1;ta co $VT\geq VP=>"="\Leftrightarrow x^2-\frac7{x^2}=x-\frac7{x^2}=0=>vo   nghiem$
2.$DK:x^2-1\geq 0;x\geq \sqrt{x^2-1}=>x\geq 1$
ta co $vt^2=(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}})^2\leq 2x\leq 2x^3(x\geq 1)<vp^2$
=>vo no

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