已知x+y=-9,xy=20,求y+1/x+1+x+1/y+1 的值

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已知x+y=-9,xy=20,求y+1/x+1+x+1/y+1 的值

已知x+y=-9,xy=20,求y+1/x+1+x+1/y+1 的值
已知x+y=-9,xy=20,求y+1/x+1+x+1/y+1 的值

已知x+y=-9,xy=20,求y+1/x+1+x+1/y+1 的值
【参考答案】
简便做法:
[(y+1)/(x+1)]+[(x+1)/(y+1)]
=[(y+1)²+(x+1)²]/[(x+1)(y+1)]
=(x²+y²+2x+2y+2)/(xy+x+y+1)
=[(x+y)²-2xy+2(x+y)+2]/[xy+(x+y)+1]
=[(-9)²-2×20+2×(-9)+2]/[20+(-9)+1]
=[81-40-16]/12
=25/12

因为x+y=-9,xy=20
所以x²+y²
=(x+y)²-2xy
=81-40
=41,
所以(y+1)/(x+1)+(x+1)/(y+1)
=(y+1)²/(x+1)(y+1)+(x+1)²/(x+1)(y+1)
=[(y+1)²+(x+1)²]/(x+1)(y+1)...

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因为x+y=-9,xy=20
所以x²+y²
=(x+y)²-2xy
=81-40
=41,
所以(y+1)/(x+1)+(x+1)/(y+1)
=(y+1)²/(x+1)(y+1)+(x+1)²/(x+1)(y+1)
=[(y+1)²+(x+1)²]/(x+1)(y+1)
=(y²+2y+1+x²+2x+1)/(xy+x+y+1)
=(41-18+2)/(20-9+1)
=25/12

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原式=【(y+1)²+(x+1)²]/(x+1)(y+1)=[(x+y)²-2xy+2(x+y)+2]/[xy+(x+y)+1]
=(81-40-18+2)/(20-9+1)=25/12

原式=[(x+1)^2+(y+1)^2]/(x+1)(y+1)
=[x^2+y^2+2(x+y)+2]/(xy+y+x+1)
=[(x+y)^2-2xy+2(x+y)+2]/(xy+x+y+1)
把X+Y=-4,XY=-12带入进去
得[(-4)^2+24-8+2]/(-15)
=-34/15
[若原式=(x+1)/(y+1)+(y+1)/(x+1)...

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原式=[(x+1)^2+(y+1)^2]/(x+1)(y+1)
=[x^2+y^2+2(x+y)+2]/(xy+y+x+1)
=[(x+y)^2-2xy+2(x+y)+2]/(xy+x+y+1)
把X+Y=-4,XY=-12带入进去
得[(-4)^2+24-8+2]/(-15)
=-34/15
[若原式=(x+1)/(y+1)+(y+1)/(x+1)]
x^2为x的平方
参考资料:http://zhidao.baidu.com/question/388351602.html

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