Question

Giải phương trình : \(x^2+\dfrac{4x^2}{x^2-4x+4}=5\)

Answer 1

\(x^2+\dfrac{4x^2}{x^2-4x+4}=5\\ x^4-4x^3+4x^2+4x^2=5x^2-20x+20\\ \Leftrightarrow x^4-4x^3+3x^2+20x-20=0\\\Leftrightarrow\left(x^4-x^3\right)-\left(3x^3-3x^2\right)+\left(20x-20\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-3x^2+20\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x+10x+20\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-5x+10\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\left(\left(x-2,5\right)^2+3,75\right) =0\)

Từ đó suy ra x=1 hoặc x=-2(Đk x khác 2)