已知x²+4y²-4x+4y+5=0求
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/27 05:11:26
![已知x²+4y²-4x+4y+5=0求](/uploads/image/z/5467408-16-8.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B%EF%BC%8B4y%26%23178%3B-4x%2B4y%2B5%3D0%E6%B1%82)
已知x²+4y²-4x+4y+5=0求
已知x²+4y²-4x+4y+5=0求
已知x²+4y²-4x+4y+5=0求
x²+4y²-4x+4y+5=0
(x-2)²+(2y+1)²=0
x=2,y=-1/2
所以
原式=(x²+y²)(x+y)(x-y)(x+2y)y²÷【(x+2y)(x-y)y(x-y)(x-y)²(x+y)²】
=(x²+y²)y÷【(x-y)(x-y)²(x+y)】
=(4+4分之1)×(-1/2)÷【5³/2³×3/2】
=-34/375
有题可知(x-2)²+(2y+1)²=0
∴x=2,y=-1/2
带入可得结果
x²+4y²-4x+4y+5=0
x²-4x+4+4y²+4y+1=0
(x-2)²+(2y+1)²=0
(x-2)²=0,(2y+1)²=0
x=2
y=-1/2
(x^4-y^4)/(x^2+xy-2y^2)*(x+2y)/(xy-y^2)÷[(-x^2-y^2)^2...
全部展开
x²+4y²-4x+4y+5=0
x²-4x+4+4y²+4y+1=0
(x-2)²+(2y+1)²=0
(x-2)²=0,(2y+1)²=0
x=2
y=-1/2
(x^4-y^4)/(x^2+xy-2y^2)*(x+2y)/(xy-y^2)÷[(-x^2-y^2)^2/y]^2
=(x^2-y^2)(x^2+y^2)/[(x+2y)(x-y)]*(x+2y)/[y(x-y)]÷[(x^2+y^2)^2/y]^2
=(x^2-y^2)(x^2+y^2)/[(x+2y)(x-y)]*(x+2y)/[y(x-y)]*y^2/(x^2+y^2)^2
=(x^2-y^2)/(x-y)*1/(x-y)*y/(x^2+y^2)
=(x-y)(x+y)/(x-y)*1/(x-y)*y/(x^2+y^2)
=(x+y)/(x-y)*y/(x^2+y^2)
=y(x+y)/[(x-y)(x^2+y^2)]
=y(x+y)/[(x-y)(x^2+y^2)]
=[-1/2*(2-1/2)]/{[2-(-1/2)][2^2+(-1/2)^2]}
=[-1/2*3/2]/{5/2*(4+1/4)}
=(-3/4)/{5/2*17/4}
=(-3/4)/(85/8)
=-3/4*8/85
=-6/85
收起