方程4x²-7x-3=0的两根为x1.x2,则x2/(x1+1) +x1/(x2+1)=?

方程4x²-7x-3=0的两根为x1.x2,则x2/(x1+1) +x1/(x2+1)=?
方程4x²-7x-3=0的两根为x1.x2,则x2/(x1+1) +x1/(x2+1)=?

方程4x²-7x-3=0的两根为x1.x2,则x2/(x1+1) +x1/(x2+1)=?
根据题意得
x1x2=-3/4 x1+x2=7/4
于是
x2/(x1+1) +x1/(x2+1)
=[x2(x2+1)+x1(x1+1)]/[(x1+1)(x2+1)]
=[x²2+x2+x²1+x1)]/(x1x2+x1+x2+1)
=[(x1+x2)²-2x1x2+(x1+x2)]/(x1x2+x1+x2+1)
=[(7/4)²-2*(-3/4)+7/4]/(-3/4+7/4+1)
=(49/16+3/2+7/4)/2
=(101/16)/2
=101/32