b)$\left\{\begin{matrix}(x+2)\sqrt{2x-1}+(y+2)\sqrt{2y-1}=2\sqrt{(x+2)(y+2)} & \\ x+y=2xy & \end{matrix}\right.$
pt (1) $(x+2)\sqrt{2x-1}+(y+2)\sqrt{2y-1}=2\sqrt{(x+2)(y+2)}\geq 2\sqrt{(x+2)(y+2)}\sqrt[4]{(2x-1)(2y-1)}$
$\Leftrightarrow 1 \geq (2x-1)(2y-1) \Leftrightarrow 0\geq 4xy-2(x+y)$
$\Leftrightarrow 0\geq 2xy-(x+y)$ Mà ta có $2xy-x-y=0$
Dấu bằng xảy ra khi $x=y=1$
c)$\left\{\begin{matrix}\frac{x^{4}}{y^{4}}+\frac{y^{4}}{x^{4}}-(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}})+\frac{x}{y}+\frac{y}{x}=-2 (1) & \\ x^{3}+x^{2}+15y+30=4\sqrt[4]{27(1-y)} \end{matrix}\right.$
Đặt $t = \frac{x}{y}+\frac{y}{x}\geq 2$
pt $(1)\Leftrightarrow (t+2)(t^3-2t^2-t+3)=0 \Leftrightarrow t=-2$
$\Leftrightarrow (x+y)^2=0 \Leftrightarrow x=-y.$
pt $(2)\Leftrightarrow x^2+x^2-15x+30=4\sqrt[4]{3.3.3(1+x)}\leq 10+x$ ( ĐK: $x+1\geq 0$)
$\Leftrightarrow (x-2)^2(x+5)\leq 0 \Leftrightarrow x=2\Rightarrow y=-2$
a) $\left\{\begin{matrix}(x^{2}+y^{2})(x+y+2)=4(y+2)(1) & \\ -(x^{2}+y^{2})=(y+2)(x+y-2) (2)& \end{matrix}\right.$
Nhân chéo (1) và (2) $\Leftrightarrow (x^2+y^2)(y+2)(x+y+2)(x+y-2)=-4(y+2)(x^2+y^2)$
$\Rightarrow y=-2$
$\Rightarrow (x+y)^2-4=-4 \Leftrightarrow (x+y)^2=0 \Leftrightarrow x=-y$.
$...$
p/s: Câu $a)$ và câu $d)$ là một .
