已知x+y=-4,xy=-12,求(y+1/x+1)+(x+1/y+1)的值.)
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已知x+y=-4,xy=-12,求(y+1/x+1)+(x+1/y+1)的值.)
已知x+y=-4,xy=-12,求(y+1/x+1)+(x+1/y+1)的值.)
已知x+y=-4,xy=-12,求(y+1/x+1)+(x+1/y+1)的值.)
x+y=-4
两边平方
x²+2xy+y²=16
所以x²+y²=40
通分
原式=(y²+2y+1+x²+2x+1)+(x+1)(y+1)
=[(x²+y²)+2(x+y)+2]/(xy+x+y+1)
=(40-8+2)/(-12-4+1)
=-34/15
由x+y=-4和xy=-12可得x和y等于2和-6(但不知谁是2谁是-6)
(y+1)/(x+1)+(x+1)/(y+1)
=[(y+1)^2+(x+1)^2]/[(x+1)(y+1)]
=(9+25)/(3*-5)
=-34/15
x+y=-4,xy=-12
∴(x+y)²=(-4)²
x²+y²+2xy=16
x²+y²=16-2xy=16-2×(-12)=40
(y+1/x+1)+(x+1/y+1)
=[(y+1)²+(x+1)²]/[(x+1)(y+1)]
=[y²+2y+1+...
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x+y=-4,xy=-12
∴(x+y)²=(-4)²
x²+y²+2xy=16
x²+y²=16-2xy=16-2×(-12)=40
(y+1/x+1)+(x+1/y+1)
=[(y+1)²+(x+1)²]/[(x+1)(y+1)]
=[y²+2y+1+x²+2x+1]/[x+y+xy+1]
=[(x²+y²+2(x+y)+2]/[(x+y)+xy+1]
=[40+2×(-4)+2]/[(-4)+(-12)+1]
=34/(-15)
=-34/15
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