Chủ đề này có 1 trả lời

[rainpoem47]

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

[Issac Newton of Ngoc Tao]

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

Ta có: $(1-y)(\sqrt{x-y}-1)+x-y-1=(x-y-1)\sqrt{y}\Leftrightarrow (x-y-1)(\frac{(1-y)}{\sqrt{x-y}+1}+1-\sqrt{y})\Leftrightarrow (x-y-1)(1-y)(\frac{1}{\sqrt{x-y}+1}+\frac{1}{1+\sqrt{y}})=0\Leftrightarrow x=y+1;y=1$

Thay y=1 suy ra x

Thay x=y+1 có:

$2y^{2}+9y+6=7\sqrt{y(y^{2}+3y+3)}\Rightarrow 2a^{2}-7ab+3b^{2}=0;a=y^{2}+3y+3,b=y;\Leftrightarrow a=3b,2a=b$

Từ đây tìm ra y suy ra x