Question

Giải phương trình :

a) \(\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

b) \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)^2\left(x+\frac{1}{x}\right)^2=\left(x-4\right)^2\)

Answer 1

\(\frac{x^2}{\left(x+2\right)}=3x^2-6x-3,x\ne-2\)

\(\Rightarrow x^2=\left(3x^2-6x-3\right)\left(x+2\right)^2\)

\(\Rightarrow x^2-\left(3x^2-6x-3\right)\left(x+2\right)^2=0\)

\(\Rightarrow x^2-\left(3x^4+12x^3+12x^2-6x^3-24x^2-24x-3x^2-12x-12\right)=0\)

\(\Rightarrow x^2-\left(3x^4+6x^3-15x^2-36x-12\right)=0\)

\(\Rightarrow16x^2-3x^4-6x^3+36x+12=0\)

\(\Rightarrow-2x^2+18x^2-3x^4-6x^3+36x+12=0\)

\(\Rightarrow-x^2\left(3x^2+6x+2\right)+\left(3x^2+6x+2\right)=0\)

\(\Rightarrow-\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\left(3x^2+6x=2\right)=0\\x^2-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{-3+\sqrt{3}}{3}\\\frac{-3-\sqrt{3}}{3},x\ne-2\\x=-\sqrt{6}\\x=\sqrt{6}\end{matrix}\right.\)

Answer 2