[tex]\left\begin\{2y^3+2x\sqrt{1-x}=3\sqrt{1-x}-y\\{y=2x^2-1+2xy\sqrt{1+x}}[/tex]
[TEX](1) \Leftrightarrow 2 (\sqrt{1-x})^3+\sqrt{1-x} = 2y^3 + y [/TEX]
[TEX] f(t) =2t^3 + t[/TEX] đồng biến [TEX]\Rightarrow (1) \Leftrightarrow \sqrt{1-x}=y[/TEX]
[TEX](2) \Leftrightarrow \sqrt{1-x} = 2x^2- 1+ 2x\sqrt{1-x^2} [/TEX]
[TEX]x= cost \ \ t \in [-\frac{\pi}{2};\frac{\pi}{2}] [/TEX]
[TEX]\sqrt{2} sin {\frac{t}{2}} = cos 2t + sin 2t [/TEX]
[TEX]\Leftrightarrow sin{\frac{t}{2}} = sin ( 2t + \frac{\pi}{4}) [/TEX]
[TEX]\Leftrightarrow \left[ \frac{t}{2} = 2t + \frac{\pi}{4} + k 2\pi \\ \frac{t}{2} = \frac{3 \pi}{4} - 2t + k 2\pi [/TEX]
[TEX]\Leftrightarrow \left[ t = -\frac{\pi}{6} \\ t = \frac{3\pi}{10} [/TEX]
[TEX]\Leftrightarrow \left[ x = cos {\frac{3\pi}{10}} \\ x = \frac{\sqrt{3}}{2}[/TEX]