Question

已知x+y=4,xy=-12 求(y+1/x+1)+(x+1/y+1)的值

Answer 1

(y+1/x+1)+(x+1/y+1)
=[(y+1)^2+(x+1)^2]/[(x+1)(y+1)]
=[x^2+2(x+y)+y^2+2]/(xy+x+y+1)
=[(x+y)^2-2xy+2(x+y)+2]/(xy+x+y+1)
因：x+y=4,xy=-12
所以原式=(16+24+8+2)/(-12+4+1)
=50/(-7)
=-50/7

Answer 2

x²+y²=(x+y)²-2xy=(-4)²-2×(-12)=40

(y+1)/(x+1)+(x+1)/(y+1)
通分得
=[(y+1)²+(x+1)²]/[(x+1)(y+1)]
展开得
=[y²+2y+1+x²+2x+1]/(xy+x+y+1)
=[(x²+y²)+2(x+y)+2]/[xy+(x+y)+1]
代入已知量
=(40+2×(-4)+2)/(-12+(-4)+1)
=34/(-15)
=-34/15

Answer 3

(y+1/x+1)+(x+1/y+1)
=[(y+1)^2+(x+1)^2]/[(x+1)(y+1)]
=[x^2+2(x+y)+y^2+2]/(xy+x+y+1)
=[(x+y)^2-2xy+2(x+y)+2]/(xy+x+y+1)
因：x+y=4,xy=-12
所以原式=(16+24+8+2)/(-12+4+1)
=50/(-7)
=-50/7

Answer 4

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