Thay x=1 vào f(x), ta được:
\(\left(m-1\right)\cdot1^2-3m\cdot1+2=0\)
\(\Leftrightarrow m-1-3m+2=0\)
\(\Leftrightarrow-2m=-1\)
hay \(m=\dfrac{1}{2}\)
Ta có :
Ta có :
\(f\left(1\right)=\left(m-1\right)1^2-3m.1+2=0\text{⇔}-2m+1=0\text{⇔}m=\dfrac{1}{2}\)
Với \(x=1\) thì \(f\left(x\right)=\left(m-1\right).1^2-3m.1+2=m-1-3m+2=-2m+1\)
\(f\left(x\right)=0\Leftrightarrow-2m+1=0\Leftrightarrow m=\dfrac{1}{2}\)
Vậy \(m=\dfrac{1}{2}\)