Theo hệ thức Vi-et ta có:
\(x_1+x_2=\frac{2\left(m-1\right)}{m+2}=\frac{2m+4}{m+2}-\frac{6}{m+2}=2-\frac{6}{m+2}\Rightarrow m+2=\frac{-6}{x_1+x_2-2}\)
\(x_1.x_2=\frac{3-m}{m+2}=\frac{-m-2}{m+2}+\frac{5}{m+2}=-1+\frac{5}{m+2}\Rightarrow m+2=\frac{5}{x_1.x_2+1}\)
Suy ra: \(-\frac{6}{x_1+x_2-2}=\frac{5}{x_1.x_2+1}\)<=>-6x1.x2-6=5x1+5x2-10 <=>5x1+5x2+6x1.x2-4 (pt cần tìm)
+ m \(\ne\)-2
\(\Delta'=\left(m-1\right)^2-\left(m+2\right)\left(3-m\right)=m^2-2m+1+m^2-m-6=2m^2-3m-5=\left(2m-5\right)\left(m+1\right)\)
\(m\ge\frac{5}{2};m\le-1\)
\(\int\limits^{x_1+x_2=\frac{2\left(m-1\right)}{\left(m+2\right)}=2-\frac{4}{m+2}}_{x_1x_2=\frac{3-m}{m+2}=-1+\frac{5}{m+2}}\)=>\(\int\limits^{5\left(S\right)=10-\frac{20}{m+2}}_{4x_1x_2=-4+\frac{20}{m+2}}\)=>5S+ 4P = 6