
已知(x-y+1)²+|x+y-2|=0,则(x-y+ 4xy/x-y)(x+y- 4xy/x+y)=
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是(x-y+ 4xy/x-y)(x+y- 4xy/x+y)?还是[x-y+ 4xy/(x-y)][x+y- 4xy/(x+y)]?
假设是后者。
解:
因为:(x-y+1)²+|x+y-2|=0
而(x-y+1)²≥0,+|x+y-2|≥0
所以:
(x-y+1)²=0…………………(1)
|x+y-2|=0……………………(2)
由(1)得:x-y+1=0…………(3)
由(2)得:x+y-2=0…………(4)
由(3)得:x=y-1……………(5)
代(5)入(4),有:y-1+y-2=0
解得:y=3/2
代入(5),有:x=3/2-1
解得:x=1/2
[x-y+4xy/(x-y)][x+y-4xy/(x+y)]
={[(x-y)^2+4xy]/(x-y)}{[(x+y)^2-4xy]/(x+y)}
=[(x^2-2xy+y^2+4xy)/(x-y)][(x^2+2xy+y^2-4xy)/(x+y)]
=[(x^2+2xy+y^2)/(x-y)][(x^2-2xy+y^2)/(x+y)]
={[(x+y)^2]/(x-y)}{[(x-y)^2]/(x+y)}
=[(x+y)^2][(x-y)^2]/[(x-y)(x+y)]
=(x+y)(x-y)
=x^2-y^2
=(1/2)^2-(3/2)^2
=1/4-9/4
=-2
假设是后者。
解:
因为:(x-y+1)²+|x+y-2|=0
而(x-y+1)²≥0,+|x+y-2|≥0
所以:
(x-y+1)²=0…………………(1)
|x+y-2|=0……………………(2)
由(1)得:x-y+1=0…………(3)
由(2)得:x+y-2=0…………(4)
由(3)得:x=y-1……………(5)
代(5)入(4),有:y-1+y-2=0
解得:y=3/2
代入(5),有:x=3/2-1
解得:x=1/2
[x-y+4xy/(x-y)][x+y-4xy/(x+y)]
={[(x-y)^2+4xy]/(x-y)}{[(x+y)^2-4xy]/(x+y)}
=[(x^2-2xy+y^2+4xy)/(x-y)][(x^2+2xy+y^2-4xy)/(x+y)]
=[(x^2+2xy+y^2)/(x-y)][(x^2-2xy+y^2)/(x+y)]
={[(x+y)^2]/(x-y)}{[(x-y)^2]/(x+y)}
=[(x+y)^2][(x-y)^2]/[(x-y)(x+y)]
=(x+y)(x-y)
=x^2-y^2
=(1/2)^2-(3/2)^2
=1/4-9/4
=-2
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(x-y+1)²+|x+y-2|=0
由非负性
x-y+1=0
x+y-2=0
解得
x=0.5
y=1.5
则
(x-y+ 4xy/x-y)(x+y- 4xy/x+y)
=(-1-4xy)(2-2xy)
=(-1-3)*(2-1.5)
=(-4)*0.5
=-2
由非负性
x-y+1=0
x+y-2=0
解得
x=0.5
y=1.5
则
(x-y+ 4xy/x-y)(x+y- 4xy/x+y)
=(-1-4xy)(2-2xy)
=(-1-3)*(2-1.5)
=(-4)*0.5
=-2
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(x-y+1)²+|x+y-2|=0,
则:
x-y+1=0
x+y-2=0
两式相加得得:
x=1/2
两式相减得:
y=3/2
(x-y+ 4xy/x-y)(x+y- 4xy/x+y)=(-1-3)(2-3/2)=-2
则:
x-y+1=0
x+y-2=0
两式相加得得:
x=1/2
两式相减得:
y=3/2
(x-y+ 4xy/x-y)(x+y- 4xy/x+y)=(-1-3)(2-3/2)=-2
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这个不难啊,平方加绝对值等于零,可知 x-y=-1 &x+y=2 求出 x=0.5 y=1.5 ,带入下式中。。欧克了。。。呵呵
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