a) \(x^2-3x+3=1\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy.............
b)Có: \(\dfrac{x^2-2x+1}{x^2+6x+9}=\dfrac{\left(x-1\right)^2}{\left(x+3\right)^2}=0\)
ĐKXĐ: \(x\ne-3\)
Pt <=> \(\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy........
a) x2 -3x +3 -1 =0
x2 -3x +2 =0 hãy nghĩ tới a+b+c =0 => x1 =1; x2 =c/a = 2
b)đây là đa thức, k phải pt
