Question

(x²-3x+3) (x²-2x+3)=2x²

Answer 1

Ta có: \(\left(x^2-3x+3\right)\left(x^2-2x+3\right)=2x^2\)

\(\Leftrightarrow\left(x^2+3\right)^2-5x\left(x^2+3\right)+6x^2-2x^2=0\)

\(\Leftrightarrow\left(x^2+3\right)^2-5x\left(x^2+3\right)+4x^2=0\)

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

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

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

mà \(x^2-x+3>0\forall x\)

nên \(x^2-4x+3=0\)

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

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: S={1;3}