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
【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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