$\begin{cases}\left ( 2y+1 \right )\sqrt{x+1}-\sqrt{\left ( x +2\right )\left ( y^{2}+1 \right )}=3y^{2}-2x+y-3 \\ \left ( 8x+10 \right )\left (2y- \sqrt{x+1}\right )=\left ( 5+\sqrt{x-1} \right )\left ( y^{2}+10\sqrt{x-1}+24\right )\end{cases}$
hum qua max . hum nay ms lm :D –  ☼SunShine❤️ 26-05-16 04:09 PM
$=> y^{2}=x 1$ –  ☼SunShine❤️ 25-05-16 08:16 PM
à , mk lại nhân kiểu khác : $Y=100=> X=99$ –  ☼SunShine❤️ 25-05-16 08:16 PM
là x 1=y^{2} bạn. –  amthambenem661 25-05-16 08:14 PM
ĐK ...
$pt(1) \Leftrightarrow  (2y+1)( \sqrt{x+1}-y) + \sqrt{x+2}(\sqrt{x+2}-\sqrt{y^{2}+1})+x+1-y^{2}=0$
$ \Leftrightarrow  (x+1-y^{2})\left[ { \frac{2y+1}{ \sqrt{x+1}+y} + \frac{\sqrt{x+2}}{ \sqrt{x+2}+ \sqrt{y^{2}+1}}+1 } \right]=0$
Vì cái :$[...] >0 \rightarrow  x+1=y^{2}$
Thế vào pt (2) :
 $ (8x+10) \sqrt{x+1}= (5+ \sqrt{x-1})(x+1+10 \sqrt{x-1}+24)$
$\Leftrightarrow  ...$ Tới đây chắc dễ r :D

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