ĐKXĐ: \(\left\{{}\begin{matrix}-1< x< 1\\x\ne0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x^2}=a\\x=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a^2+b^2=1\\\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{5}{12}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=1\\ab=\dfrac{12}{5}\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^2-\dfrac{24}{5}\left(a+b\right)-1=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b=5\left(loại\right)\\a+b=-\dfrac{1}{5}\end{matrix}\right.\) (loại do \(\left(a+b\right)^2\le2\left(a^2+b^2\right)=2\))
\(\Rightarrow x+\sqrt{1-x^2}=-\dfrac{1}{5}\)
\(\Rightarrow\sqrt{1-x^2}=-x-\dfrac{1}{5}\left(x\le-\dfrac{1}{5}\right)\)
\(\Rightarrow1-x^2=x^2+\dfrac{2x}{5}+\dfrac{1}{25}\left(x\le-\dfrac{1}{5}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(loại\right)\\x=-\dfrac{4}{5}\end{matrix}\right.\)
đề là \(\dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{e}=\dfrac{5}{1z}\) à chị
