帮我解一下分式计算已知x+y=-4,xy=-12,求(y+1）/（x+1) +（x+1)/(y+1)的值

帮我解一下分式计算已知x+y=-4,xy=-12,求(y+1）/（x+1) +（x+1)/(y+1)的值
帮我解一下分式计算
已知x+y=-4,xy=-12,求(y+1）/（x+1) +（x+1)/(y+1)的值

帮我解一下分式计算已知x+y=-4,xy=-12,求(y+1）/（x+1) +（x+1)/(y+1)的值
x+y=-4 (x+y)^2=16 xy=-12 x^2+y^2=40
(y+1）/（x+1) +（x+1)/(y+1)
={(y+1)^2+(x+1)^2}/(x+1)(y+1)
=(y^2+2y+1+x^2+2x+1)/(xy+x+y+1)
=-34/15

(y+1)/(x+1)+(x+1)/(y+1)通分
=[(y+1)(y+1)+(x+1)(x+1)]/[(x+1)(y+1)]
=(y^2+2y+1+x^2+2x+1)/(xy+x+y+1)，其中,^2表示平方
=[x^2+y^2+2(x+y)+2]/[xy+(x+y)+1]
=[(x+y)^2-2xy+2(x+y)+2]/[xy+(x+y)+1]
=[(-4)^2-2×(-12)+2(-4)+2]/[-12-4+1]
=[16+24-8+2]/[-15]
=-34/15