Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=\frac32;x_1x_2=\frac{c}{a}=-\frac12\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(\frac32\right)^2-2\cdot\frac{-1}{2}=\frac94+1=\frac{13}{4}\)
\(\left(x_2+1\right)\left(x_1+1\right)=x_1x_2+\left(x_1+x_2\right)+1\)
\(=-\frac12+\frac32+1=1+1=2\)
\(A=\frac{x_1-1}{x_2+1}+\frac{x_2-1}{x_1+1}\)
\(=\frac{\left(x_1-1\right)\left(x_1+1\right)+\left(x_2-1\right)\left(x_2+1\right)}{\left(x_2+1\right)\left(x_1+1\right)}\)
\(=\frac{x_1^2-1+x_2^2-1}{\left(x_1+1\right)\left(x_2+1\right)}=\frac{x_1^2+x_2^2-2}{\left(x_1+1\right)\left(x_2+1\right)}\)
\(=\frac{\frac{13}{4}-2}{2}=\frac54:2=\frac58\)
