ĐKXĐ:..
Đặt \(2\sqrt{1-x}-\sqrt{x+1}=t\Rightarrow t^2=4\left(1-x\right)-4\sqrt{1-x^2}+x+1\)
\(\Rightarrow t^2=4-4x-4\sqrt{1-x^2}+x+1\Rightarrow\sqrt{1-x^2}=\frac{-t^2-3x+5}{4}\)
\(\Rightarrow pt\Leftrightarrow t+\frac{3}{4}.\left(5-t^2-3x\right)=3-x\)
\(\Leftrightarrow t+\frac{15}{4}-\frac{3}{4}t^2-\frac{9}{4}x=3-x\)
\(\Leftrightarrow\frac{3}{4}t^2-t+\frac{5}{4}x-\frac{3}{4}=0\)
Đặt \(\frac{5}{4}x-\frac{3}{4}=k\Rightarrow\Delta=1-4.\frac{3}{4}k=1-3k\ge0\Leftrightarrow k\le\frac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{1+\sqrt{1-3k}}{\frac{3}{2}}\\t=\frac{1-\sqrt{1-3k}}{\frac{3}{2}}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}t=\frac{2+2\sqrt{1-3\left(\frac{5}{4}x-\frac{3}{4}\right)}}{3}\\t=\frac{2-2\sqrt{1-3\left(\frac{5}{4}x-\frac{3}{4}\right)}}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{2+2\sqrt{\frac{13}{4}-\frac{15}{4}x}}{3}\\t=\frac{2-2\sqrt{\frac{13}{4}-\frac{15}{4}x}}{3}\end{matrix}\right.\)
Thay t bằng cái bên trên đầu bài rồi giải nốt nha cậu. Hình như hơi dài xíu :)
\(2\sqrt{1-x}-\sqrt{1+x}+3\sqrt{1-x^2}=3-x\) (đk: \(-1\le x\le1\))
<=>\(2\left(1-x\right)+x+1-2\sqrt{1-x}+\sqrt{1+x}-3\sqrt{1-x^2}=0\)
<=>\(2\sqrt{1-x}\left(\sqrt{1-x}-1-\sqrt{x+1}\right)-\sqrt{x+1}\left(\sqrt{1-x}-\sqrt{x+1}-1\right)=0\)
<=>\(\left(2\sqrt{1-x}-\sqrt{x+1}\right)\left(\sqrt{1-x}-1-\sqrt{x+1}\right)=0\)
<=>\(\left[{}\begin{matrix}2\sqrt{1-x}=\sqrt{x+1}\left(1\right)\\\sqrt{1-x}=1+\sqrt{x+1}\left(2\right)\end{matrix}\right.\)
pt (1) <=>\(4\left(1-x\right)=x+1\)<=> \(4-4x=x+1\) <=> \(x=\frac{3}{5}\)(t/m)
pt (2)<=>\(1-x=1+x+1+2\sqrt{x+1}\)
<=> \(-2x-1=2\sqrt{x+1}\)( đk: \(-1\le x\le\frac{1}{2}\))
<=> \(4x^2+4x+1=4x+4\) <=> \(4x^2=3\) <=> \(\left[{}\begin{matrix}x=\frac{\sqrt{3}}{2}\left(l\right)\\x=-\frac{\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt (*) có tập nghiệm \(S=\left\{-\frac{\sqrt{3}}{2},\frac{3}{5}\right\}\)
