a.
`x^2-2(m-1)x+m+1=0` `(1)`
`Δ'=(m-1)^2-m-1=m^2-2m+1-m-1=m^2-3m`
Để phương trình `(1)` có 2 nghiệm phân biệt thì:
`Δ'>0`
`⇔ m^2-3m>0`
`⇔` \(\left[ \begin{array}{l}m>3\\m<0\end{array} \right.\)
b.
Theo Viét ta có: $\begin{cases}x_1+x_2=2m-2\\x_1x_2=m+1\end{cases}$
Ta có: $\begin{cases}x_1=3x_2&\\ x_1+x_2=2m-2\\ x_1x_2=m+1\end{cases}$
`⇔` $\begin{cases}4x_1=2m-2&\\ x_1=3x_2\\ x_1x_2=m+1\end{cases}$
`⇔` $\begin{cases}x_1=\frac{m-1}{2}&\\ x_2=\frac{3m-3}{2}\end{cases}$
`x_1x_2=m+1`
`⇔` `(\frac{m-1}{2}). 3/2(m-1)=m+1`
`⇔` `3/4(m^2-2m+1)=m+1`
`⇔` `3/4m^2-5/2m-1/4=0`
`⇔` \(\left[ \begin{array}{l}m=\frac{5+2\sqrt[]{7}}{3} (n)\\m=\frac{5-2\sqrt[]{7}}{3}(l)\end{array} \right.\)
`⇔` `m=\frac{5+2\sqrt[7]}{3}`
