若x^2-2xy+y^2=0(且x≠0),求(2x+3y)/x+y
若x^2-2xy+y^2=0(且x≠0),求(2x+3y)/x+y
解:
因为 x^2-2xy+y^2=0 (且x≠0),求(2x+3y)/x+y
所以 (x-y)^2=0
所以 x=y≠0
所以 (2x+3y)/(x+y)=(2y+3y)/(y+y)=5/2
O(∩_∩)O~
x²-2xy+y²=0 则(x-y)²=0 解得:x=y 所以所求式子:
(2x+3y)/x+y=(5x)/x+x=5+x
如果分母是(x+y),则所求式子:(2x+3y)/(x+y)=(5x)/(2x)=5/2=2.5
回答完毕保证正确
已知:x²-2xy+y²=0,
(x-y)²=0,
x-y=0,
x=y。
已知:x≠0,
(2x+3y)/x+y
=(2x+3x)/x+x
=5x/x+x
=5+x。
【如果】
(2x+3y)/(x+y)
=(2x+3x)/(x+x)
=5x/(2x)
=5/2
x^2-2xy+y^2=0
(x-y)^2=0
x-y=0
x=y
(2x+3y)/x+y
=(2y+3y)/y+y
=5y/y+y
=5+y
如果是求(2x+3y)/(x+y)的话
=(2x+3y)/(x+y)
=(2y+3y)/(y+y)
=5y/2y
=5/2
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