作业帮已知x+y=-4.xy=12,求x+1分之y+1加上y+1分之x+1的值
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已知x+y=-4.xy=12,求x+1分之y+1加上y+1分之x+1的值
已知x+y=-4.xy=12,求x+1分之y+1加上y+1分之x+1的值
已知x+y=-4.xy=12,求x+1分之y+1加上y+1分之x+1的值
(y+1)/(x+1)+(x+1)/(y+1)
=[(y+1)²+(x+1)²]/[(x+1)(y+1)]
=(y²+2y+1+x²+2x+1)/(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)
=-14/9
(y+1)/(x+1) + (x+1)/(y+1)
=(y+1)^2+(x+1)^2 /(x+1)(y+1)
=(x^2+y^2+2x+2y+2)/(xy+x+y+1)
已知x+y=4.xy=-12把？不然原式无解！
一式平方得x^2+y^2+2xy=16
减2倍的2式得x^2+y^2=40
原式=-14
通分之后很简单啊
没算错是9分之14
可以自己再看看
(y+1) / (x+1) + (x+1) / (y+1)
= {(y+1)^2+(x+1)^2} /{(x+1)(y+1)}
= {x^2+2x+1+y^2+2y+1} / {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}
= -14/9
=(y+1)/(x+1)+(x+1)/(y+1)
=[(y+1)²+(x+1)²]/(x+1)(y+1)
=(y²+2y+1+x²+2x+1)/(xy+x+y+1)
=[(x+y)²-2xy+2(x+y)+2]/(xy+x+y+1)
把x+y=-4.xy=12代人上式得：
[(-4)²-2×12+2(-4)+2]/(12-4+1)
=-14/9
[（y+1)/(x+1)]+[(x+1)/(y+1)]
=[(y+1)²+(x+1)²]/(x+1)(y+1)
=[x²+y²+2（x+y)+2]/(xy+x+y+1)
=[(x+y)²-2xy+8+2]/(xy+x+y+1)
=(16-24+10)/(12-4+1)
=2/9
原式=(x+1)＾+(y+1)＾/(x+1)(y+1)
=x＾+1+2x+y＾+1+2y/xy+x+y+1
=x＾+y＾+3(x+y)+xy+3
因为x+y=-4
xy=12
所以x＾+y＾=(x+y)＾－2xy
=-8
所以原式=-8+3×(-4)+12+3
=-5
注：＾为“平方”
负9分之14，你先列出待解的式子，通分，合并成（x+y）和xy形式的，代入各自实际值就行