[TEX]
[TEX]64. \\ x^3 + \sqrt{(1-x^2)^3} = x\sqrt{2.(1-x^2)}[/TEX]
[TEX]DK: 0 \le x \le 1 [/TEX]
[TEX]x = siny ( y \in [ - \frac{\pi}{2} ; \frac{\pi}{2} ] )[/TEX]
[TEX](pt) \Leftrightarrow sin^3 y + cos^3 y = \sqrt{2} sin y . cos y [/TEX]
[TEX]t = sin y + cos y ( |t| \le \sqrt{2} ) \Rightarrow t^3 - 3t ( \frac{t^2 -1}{2} ) = \sqrt{2} \frac{t^2-1}{2} \\ \Leftrightarrow 2t^3 - 3t( t^2 -1) = \sqrt{2} ( t^2 -1) \\ \Leftrightarrow t^3 + \sqrt{2} t^2- 3t - \sqrt{2} = 0 \\ \Leftrightarrow ( t - \sqrt{2} ) ( t+1+\sqrt{2})(t+\sqrt{2}-1)=0 [/TEX]
[TEX]\Leftrightarrow \left[ t = sqrt{2} (thoa) \\ t = -1 - \sqrt{2} (loai) \\ t =1 - \sqrt{2} (thoa) [/TEX]
[TEX]TH1: t = \sqrt{2} \Leftrightarrow sin( y + \frac{\pi}{4} ) = 1 \Leftrightarrow y = \frac{\pi}{4} \Leftrightarrow x = \frac{\sqrt{2}}{2} [/TEX]
[TEX]TH2 : t = 1 - \sqrt{2} \Leftrightarrow x + \sqrt{1-x^2 } = 1 - \sqrt{2} [/TEX]
Đặt : [TEX]\left{ u = x \\ v= \sqrt{1-x^2} ( v \ge0 ) [/TEX]. Ta có :
[TEX]\left{ u+v = 1 - \sqrt{2} \\ u^2 + v^2 = 1 \right. \Leftrightarrow \left{ u+v= 1- \sqrt{2} \\ uv =1-\sqrt{2} [/TEX].
u,v là nghiệm của phương trình :
[TEX]X^2 -( 1 - \sqrt{2} ) X + ( 1- \sqrt{2} ) = 0 [/TEX]
[TEX]\Leftrightarrow \left[ X = \frac12 ( 1 - \sqrt{2} - \sqrt{2\sqrt{2}-1} ) \\ X = \frac12 ( 1 - \sqrt{2} + \sqrt{2\sqrt{2}-1} )[/TEX]
[TEX]\Rightarrow \left{ u = \frac12 ( 1 - \sqrt{2} - \sqrt{2\sqrt{2}-1} ) \\ v = \frac12 ( 1 - \sqrt{2} + \sqrt{2\sqrt{2}-1} ) \right. ( do v \ge 0) [/TEX]
[TEX]\Leftrightarrow x = \frac12 ( 1 - \sqrt{2} - \sqrt{2\sqrt{2}-1} ) [/TEX]
Kết luận :
[TEX]\left[ x = \frac{\sqrt{2}}{2} \\ x = \frac12 ( 1 - \sqrt{2} - \sqrt{2\sqrt{2}-1} ) [/TEX]