Đk "; x<=+-1
\(h=\sqrt{1-x^2}\)
=<> \(x=\sqrt{1-h^2}\)
\(\dfrac{1}{h^2}=\dfrac{3\sqrt{1-h^2}}{h}-1\)
\(\dfrac{1}{h^2}=\dfrac{3\sqrt{1-h^2}-h}{h}\)
\(1=3h\sqrt{1-h^2}-h^2\)
\(\left(1+h^2\right)^2=9h^2\left(1-h^2\right)\)
\(1+2h^2+h^4=9h^2-9h^4\)
\(10h^4-7h^2+1=0\)
\(h^2=\dfrac{1}{2}\)<=> \(1-x^2=\dfrac{1}{2}\)<=>\(x=+-\sqrt{\dfrac{1}{2}}\)
\(h^2=\dfrac{1}{5}\)<=>\(1-x^2=\dfrac{1}{5}\)<=>\(x=+-\dfrac{2}{\sqrt{5}}\)
vậy cái này sai ở đâu ak??
+ ĐK: \(1-x^2\ge0\Leftrightarrow x^2< 1\)
+ pt đã cho \(\Leftrightarrow\dfrac{1}{1-x^2}=\dfrac{3x}{\sqrt{1-x^2}}-1\Leftrightarrow\sqrt{1-x^2}=3x\sqrt{1-x^2}-\left(1-x^2\right)\sqrt{1-x^2}\Leftrightarrow\left(1-x^2-3x+1\right)\sqrt{1-x^2}=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x^2}=0\left(L\right)\\1-x^2-3x+1=0\left(2\right)\end{matrix}\right.\)
(2) \(\Leftrightarrow-x^2-3x+2=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{17}}{2}\left(N\right)\\x=\dfrac{-3-\sqrt{17}}{2}\left(L\right)\end{matrix}\right.\)
KL: sao nghiệm tào lao thế!!!??!! T_T!!!
