Question

giải pt theo toán 9: \(x^2\)-3\(x^{ }\)+2=0

Answer 1

Ta có a+b+c=\(1+(-3)+2=0\)

=>  \(x=1\\ x=\dfrac{c}{a}=\dfrac{2}{1}=2\)

Vậy pt có 2 no \(x=1\\ x=2\)

Answer 2

\(x^2-3x+2=0.\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0.\\x-1=0.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)