\(\text{Để}\)\(f\left(x\right)\)\(\text{có nghiệm là}\)\(x=-1\)

\(m\left(-1\right)^2-2\left(-1\right)+8=0\)

\(\Leftrightarrow m+2+8=0\)

\(\Leftrightarrow m+10=0\)

\(\Leftrightarrow m=-10\)

\(\text{Vậy}\)\(m=-10\)\(\text{thì}\)\(f\left(x\right)\)\(\text{có nghiệm là}\) \(-1\)

a.\(x^2-3x=x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

c.\(x^3+x^2+x+1=x^2\left(x+1\right)+\left(x+1\right)=\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\text{ ( vô lí )}\\x=-1\end{cases}}}\Leftrightarrow x=-1\)

d.\(x^2+\sqrt{2015}=0\Leftrightarrow x^2=-\sqrt{2015}< 0\text{ (vô nghiệm)}\)

mk ko bt bạn ah !!

(x+2)*(2x-1)=0

TH1: x+2=0              TH2: 2x-1=0

        x=-2                          2x=1

                                            x=1/2

vậy x= -2 hoặc 1/2