Question

giải pt :
(x²-3x+3)(x²-2x+3)=2x²

Answer 1

Answer 2

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

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

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

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

\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-x+3=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)