- 八年级数学化简求值十万火急已知x^2+4y^2+x^2y^2
- 已知x^2+4y^2+x^2y^2-6xy+1=0 求
x^4-y^4/2x-y . 2xy-y^2/xy-y^2 . (xy/x^2+y^2)^2 .
1/x+y 的值?
要详细过程 谢谢详细 急急急 十万火急
- (x^4-y^4)/(2x-y)*(2xy-y^2)/(xy-y^2)*(xy/x^2+y^2)^2*(1/x+y)
=[(x^2+y^2)*(x+y)*(x-y)/(2x-y)]*[y(2x-y)/y(x-y)]*[x^2*y^2/(x^2+y^2)^2]*[1/(x+y)]
=x^2*y^2/(x^2+y^2)………………………………………………(1)
已知x^2+4y^2+x^2y^2-6xy+1=0
===> (x^2-4xy+4y^2)+(x^2y^2-2xy+1)=0
===> (x-2y)^2+(xy-1)^2=0
因为(x-2y)^2≥0,且(xy-1)^2≥0
那么,要满足它们两者之和为零,则必须是两个同时为零
即:x=2y,且xy=1
代入(1)得到:原式=(2y)^2*y^2/[(2y)^2+y^2]=4y^4/(5y^2)=(4/5)y^2
又由x=2y代入xy=1得到:2y*y=1
所以,y^2=1/2
所以,原式=(4/5)*(1/2)=2/5.