Đáp án:
`\sqrt{(1+x)^3}-\sqrt{(1-x)^3}=2+\sqrt{1-x^2}`
Điều kiện:`1+x>=0,1-x>=0,1-x^2>=0`
`<=>-1<=x<=1`.
`pt<=>(\sqrt{1+x}-\sqrt{1-x})(1+x+\sqrt{(1-x)(1+x)}+1-x)=2+\sqrt{1-x^2}`
`<=>(\sqrt{1+x}-\sqrt{1-x})(2+\sqrt{1-x^2})=2+\sqrt{1-x^2}`
`<=>(2+\sqrt{1-x^2})(\sqrt{1+x}-\sqrt{1-x}-1)=0`
Vì `2+\sqrt{1-x^2}>=2>0`
`<=>\sqrt{1+x}-\sqrt{1-x}-1=0`
`<=>\sqrt{1+x}=\sqrt{1-x}+1`
`<=>1+x=1-x+1+2\sqrt{1-x}`
`<=>1+x=2-x+2\sqrt{1-x}`
`<=>2x-1=2\sqrt{1-x}`
`ĐK:2x-1>=0<=>x>=1/2`
Kết hợp điều kiện:`1/2<=x<=1`
`<=>4x^2-4x+1=4(1-x)`
`<=>4x^2-4x+1=4-4x`
`<=>4x^2=3`
`<=>x^2=3/4`
`<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt3}{2}(tm)\\x=-\dfrac{\sqrt3}{2}(l)\end{array} \right.\)
Vậy pt có nghiệm duy nhất `x=\sqrt3/2`
