a,Đk: \(\left\{{}\begin{matrix}\sqrt{1+x}\ge0\\\sqrt{1-x}\ge0\\x\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}1+x\ge0\\1-x\ge0\\x\ne0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-1\\x\le1\\x\ne0\end{matrix}\right.\) <=> \(-1\le x\le1,x\ne0\)
b, A= \(\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}-\frac{1-x}{\sqrt{1-x^2}-1+x}\right)\left(\sqrt{\frac{1}{x^2}-1}-\frac{1-x}{x}\right).\frac{x}{1-x+\sqrt{1-x^2}}\)
=\(\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}-\frac{\left(\sqrt{1-x}\right)^2}{\sqrt{1-x}\left(\sqrt{1+x}-\sqrt{1-x}\right)}\right)\left(\sqrt{\frac{1-x^2}{x^2}}-\frac{1-x}{x}\right).\frac{x}{\left(\sqrt{1-x}\right)^2+\sqrt{\left(1-x\right)\left(1+x\right)}}\)
=\(\left(\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}-\frac{\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1-x^2}}{\left|x\right|}-\frac{1-x}{x}\right).\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)( do x>0) (1)
Tại \(0\le x\le1\) => \(\left|x\right|=x\)
Từ (1) <=> A=\(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.\frac{\sqrt{\left(1-x\right)\left(1+x\right)}-\left(1-x\right)}{x}.\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)
=\(\frac{\sqrt{1-x}\left(\sqrt{1+x}-\sqrt{1-x}\right)}{x}.\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)
=\(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1-x}+\sqrt{1+x}}\)=\(\frac{\left(\sqrt{1+x}-\sqrt{1-x}\right)^2}{\left(\sqrt{1-x}+\sqrt{1+x}\right)\left(\sqrt{1+x}-\sqrt{1-x}\right)}=\frac{1+x-2\sqrt{\left(1+x\right)\left(1-x\right)}+1-x}{1+x-\left(1-x\right)}=\frac{2-2\sqrt{1-x^2}}{2x}=\frac{1-\sqrt{1-x^2}}{x}\)
Tại \(-1\le x< 0\)
Từ (1) <=> \(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}.\left(\frac{-\sqrt{1-x^2}}{x}-\frac{1-x}{x}\right).\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)
=\(\frac{-\sqrt{\left(1-x\right)\left(1+x\right)}-\left(\sqrt{1-x}\right)^2}{x}.\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)
=\(\frac{-\sqrt{1-x}\left(\sqrt{1+x}+\sqrt{1-x}\right)}{x}.\frac{x}{\sqrt{1-x}\left(\sqrt{1-x}+\sqrt{1+x}\right)}\)
=-1
Vậy \(\left\{{}\begin{matrix}0\le x\le1\\-1\le x< 0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}A=\frac{1-\sqrt{1-x^2}}{x}\\A=-1\end{matrix}\right.\)
Lê Thị Thục Hiền
Nguyễn Việt Lâm
