a: Sửa đề: \(x^2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)-6=0\)(1)
ĐKXĐ: x<>0
(1)=>\(\left(x+\dfrac{1}{x}\right)^2-2-2\left(x+\dfrac{1}{x}\right)-6=0\)
=>\(\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-8=0\)
=>\(\left(x+\dfrac{1}{x}-4\right)\left(x+\dfrac{1}{x}+2\right)=0\)
=>\(\dfrac{x^2+1-4x}{x}\cdot\dfrac{x^2+1+2x}{x}=0\)
=>\(\left(x^2-4x+1\right)\left(x^2+2x+1\right)=0\)
=>\(\left[{}\begin{matrix}x^2-4x+1=0\\x^2+2x+1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=2\pm\sqrt{3}\left(nhận\right)\end{matrix}\right.\)
b: \(x^2+2x=\sqrt{x^2+2x+1}+5\)
=>\(x^2+2x+1=\left|x+1\right|+6\)
=>\(\left(\left|x+1\right|\right)^2-\left|x+1\right|-6=0\)
=>\(\left(\left|x+1\right|-3\right)\left(\left|x+1\right|+2\right)=0\)
=>\(\left|x+1\right|-3=0\)
=>|x+1|=3
=>\(\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
