
计算:(1)x-y/2x+y/2x (2)5/x-5+5/5-x (3)(5x/x²-2xy+y² )-(5y/x²-2xy+y²)?
2020-08-25 · 知道合伙人教育行家
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(1)
(x-y)/(2x)+y/(2x)=(x-y+y)/(2x)=2x/(2x)=1
(2)
5/(x-5)+5/(5-x)= 5/(x-5)-5/(x-5) = 0
(3)
5x/(x²-2xy+y² )-5y/(x²-2xy+y²)=(5x-5y)/(x²-2xy+y²)=5(x-y)/(x-y)²=5/(x-y)
(4)
x²/(x²-y²)-2xy/(x²-y²)+y²/(x²-y²)=(x²-2xy+y²)/(x²-y²)=(x-y)²/[(x+y)(x-y)]=(x-y)/(x+y)
(x-y)/(2x)+y/(2x)=(x-y+y)/(2x)=2x/(2x)=1
(2)
5/(x-5)+5/(5-x)= 5/(x-5)-5/(x-5) = 0
(3)
5x/(x²-2xy+y² )-5y/(x²-2xy+y²)=(5x-5y)/(x²-2xy+y²)=5(x-y)/(x-y)²=5/(x-y)
(4)
x²/(x²-y²)-2xy/(x²-y²)+y²/(x²-y²)=(x²-2xy+y²)/(x²-y²)=(x-y)²/[(x+y)(x-y)]=(x-y)/(x+y)
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