已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 14:11:31
已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值
xQJ0} Y'l(})zܕ2d 7˰3/ۅ2KU_L}prb.y<C-1̸*i6"TVs´ SMAj7%kSVCʲ[ϒ8u#@4q;!LAf,qTR_ -ϡ)҇m əkd DNU̒ KѸH TO坜6Ci_÷ty>u4*7HyFxYe2De:

已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值
已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值

已知x^2+4y^2-2x-2x+8y+5=0,求x^2-y^2/2x^2+xy-y^2*2y-y/xy-y/(x^2+y^2/y)^2的值
x²+4y²-2x+8y+5=0
(x-1)²+(2y+2)²=0
得 x=1,y=-1
则 x²-y²/2x²+xy-y²*2y-y/xy-y/(x²+y²/y)²
=1-1/2-1+2-1+1/4
=1/2+1/4
=3/4

原题是多打了一个2x吗?。。如果没有那个2x,那么原式可以化为:(x-1)^2+4(y+1)^2=0,然后就x=1 y=-1.