已知x,y为有理数,且x^2+2y+y^2-4x+5=0,求(x+y)(x-y)的值
x^2+2y+y^2-4x+5=0 除以 y^2
1/y^2 *(x^2-4x+5)+1/y *2 +1 = 0
x,y为有理数 b^2-4ac =0
2^2-4*(x^2-4x+5)*1=0
x^2-4x+4=0
(x-2)^2=0
x=2
x^2+2y+y^2-4x+5=0
4+2y+y^2-8+5=0
y^2+2y+1=0
(y+1)^2=0
y=-1
(x+y)(x-y)=(2-1)(2+1)=3
。 全部
x^2+2y+y^2-4x+5=0 除以 y^2
1/y^2 *(x^2-4x+5)+1/y *2 +1 = 0
x,y为有理数 b^2-4ac =0
2^2-4*(x^2-4x+5)*1=0
x^2-4x+4=0
(x-2)^2=0
x=2
x^2+2y+y^2-4x+5=0
4+2y+y^2-8+5=0
y^2+2y+1=0
(y+1)^2=0
y=-1
(x+y)(x-y)=(2-1)(2+1)=3
。
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