[TEX]\left\{ \begin{array}{l} x\sqrt[]{y}+2y\sqrt[]{x}=3x\sqrt[]{2x-1} \\ y\sqrt[]{x}+2x\sqrt[]{y}=3y\sqrt[]{2y-1} \end{array} \right[/TEX] Điều kiện: $x,y$ \geq $\frac{1}{2}$
\Rightarrow $x\sqrt[]{y}+2y\sqrt[]{x}-y\sqrt[]{x}+2x\sqrt[]{y}=3x\sqrt[]{2x-1}-3y\sqrt[]{2y-1}$
\Leftrightarrow $\sqrt[]{x^2y}-\sqrt[]{xy^2}+3\sqrt[]{2x^3-x^2}-3\sqrt[]{2y^3-y^2}$ (vì $x,y$ \geq $\frac{1}{2}$)
\Leftrightarrow $\frac{x^2y-xy^2}{\sqrt[]{x^2y}+\sqrt[]{xy^2}}+3.\frac{2x^3-x^2-2y^3+y^2}{\sqrt[]{2x^3-x^2}+\sqrt[]{2y^3-y^2}}=0$
\Leftrightarrow $\frac{xy(x-y)}{\sqrt[]{x^2y}+\sqrt[]{xy^2}}+\frac{(x-y)(2x^2+y^2+2xy-x-y)}{\sqrt[]{2x^3-x^2}+\sqrt[]{2y^3-y^2}}=0$
\Leftrightarrow $(x-y)(\frac{xy}{\sqrt[]{x^2y}+\sqrt[]{xy^2}}+3.\frac{(x+y)(x+y-1)+x^2+y^2}{\sqrt[]{2x^3-x^2}+\sqrt[]{2y^3-y^2}})=0$
\Leftrightarrow $x=y$ do $\frac{xy}{\sqrt[]{x^2y}+\sqrt[]{xy^2}}+3.\frac{(x+y)(x+y-1)+x^2+y^2}{\sqrt[]{2x^3-x^2}+\sqrt[]{2y^3-y^2}}$> 0 d $x,y$ \geq $\frac{1}{2}$
\Rightarrow $3x\sqrt[]{x}=3x\sqrt[]{2x-1}$
\Rightarrow $x=y=1$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
