Giải hệ phương trình ?

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xuanquynh97

ĐK: $\begin{cases} x^2-x-y ≥ 0 &\\
x ≥ \dfrac{1}{2} &\\
x \not=y &
\end{cases}$

PT (1) \Leftrightarrow $\sqrt{x^2-x-y}(\sqrt[3]{x-y}-1)+(\sqrt{x^2-x-y}-y)=0$

\Leftrightarrow $$\sqrt{x^2-x-y}.\dfrac{x-y-1}{\sqrt[3]{(x-y)^2}+\sqrt[3]{x-y}+1}+\dfrac{x^2-x-y-y^2}{\sqrt{x^2-x-y}+y}=0$$

\Leftrightarrow $$\sqrt{x^2-x-y}.\dfrac{x-y-1}{\sqrt[3]{(x-y)^2}+\sqrt[3]{x-y}+1}+\dfrac{(x-y-1)(x+y)}{\sqrt{x^2-x-y}+y}=0$$

\Leftrightarrow $$(x-y-1)(\dfrac{\sqrt{x^2-x-y}}{\sqrt[3]{(x-y)^2}+\sqrt[3]{x-y}+1}+\dfrac{x+y}{\sqrt{x^2-x-y}+y})=0$$

\Leftrightarrow $y=x-1$

Thay vào PT (2) ta được

$4x^2-4x+2-3\sqrt{2x-1}=11$

\Leftrightarrow $(2x-1)^2-3\sqrt{2x-1}-10=0$

\Leftrightarrow $\sqrt{2x-1}=2$

\Leftrightarrow $x=\dfrac{5}{2}$

\Rightarrow $y=\dfrac{3}{2}$
 
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