giải hệ:
$\left\{\begin{matrix} & \sqrt{x+y+1}+1=4(x+y)^2+\sqrt{3(x+y)}\\ & x^3+5x^2+6x=(3x+2y+1)(\sqrt{3-2y}+\sqrt{5-x})\end{matrix}\right.$

Bài khai tuần ^^ 

ĐK : $y  \leq  \frac{3}{2}; x \leq  5; (x+y) \geq  0$
$pt(1) \Leftrightarrow  \frac{x+y+1-3x-3y}{\sqrt{x+y+1}+\sqrt{3(x+y)}}+(1-2x-2y)(1+2x+2y)=0$
$\Leftrightarrow  (1-2x-2y)\left[ { \frac{1}{\sqrt{x+y+1}+\sqrt{3(x+y)}}+2x+2y+1} \right]=0$
Mà cái  $[...] >0$
$\rightarrow  2y=1 -2x$
Thế vào pt(2) , ta được : 
 $x^{3}+5x^{2}+6x=(3x+1-2x+1)(\sqrt{3-1+2x}+\sqrt{5-x})$
$\Leftrightarrow  ...$
thế hả :v dễ khỏi lm đi , mà cmt sao k vote đi -_- đồ ki bo –  ☼SunShine❤️ 29-05-16 08:46 AM
giải nốt phần còn lại đi... dễ mà –  tran85295 29-05-16 08:43 AM

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