Question

已知x^2+4y^2+x^2y^2+1=6xy.求:(x^4-y^4/(2x^2+xy-y^2)*(2x-y)/(xy-y^2)再初以[(x^2+y^2)/y]^2

再初以=初号 详细过程!!!
那个故y= 什么? 看不懂

Answer 1

x^2+4y^2+x^2y^2+1=6xy
即：x^2-4xy+y^4+x^2y^2-2xy+1=0
亦即：(x-2y)^2+(xy-1)^2=0
所以x-2y=xy-1=0
即有x=2y，y^2=1/2

则[(x^4-y^4)/(2x^2+xy-y^2)]*[(2x-y)/(xy-y^2)]/[(x^2+y^2)/y]^2
=[(16y^2-y^4)/(8y^2+2y^2-y^2)]*[(4y-y)/(2y^2-y^2)]/[(4y^2+y^2)/y]^2
=(15y^4/9y^2)*(3y/y^2)^2/(5y)^2
=(5/3)*y^2*(9/y^2)/25y^2
=(3/5)y^2
=3/10

Answer 2

巳知是：(x-2y)的平方+(xy-1)的平方=0，利用非负数的性质

得x=2y ，且xy=1

故y= ±√2/2

原式化简后是:y/(x的平方+y的平方)= 1/5y=±√2/5