`a, \Delta'=b'^2-ac=(-(m-3))^2-(2m-7)=m^2-6m+9-2m+7=m^2-8m+16=(m-4)^2ge0` với mọi `m`
`b,` Áp dụng hệ thức Viet ta có:
`1/(x_1+1)+1/(x_2+1)=m`
`<=>(x_2+1)/((x_1+1)(x_2+1))+(x_1+1)/((x_1+1)(x_2+1))=m`
`<=>(x_1+x_2+2)/(x_1x_2+x_1+x_2+1)=m`
`<=>(2(m-3)+2)/(2m+7-2(m-3)-1)=m`
`<=>(m-2)/6=m`
`<=>(m-2)/6=(6m)/6`
`<=>m-2=6m`
`<=>m=-2/5`
Vậy `m=-2/5`
$#Blink$ $\boxed{\text{@Rosé}}$