2道哦

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2道哦

2道哦
2道哦



2道哦
[(x-y)/(2x+1)]*[(2x+1)/(x+y)-(2xy+y)/(y^2-x^2)]+y/(x+y)
=[(x-y)/(2x+1)]*[(2x+1)/(x+y)+(2xy+y)/(x^2-y^2)]+y/(x+y)
=[(x-y)/(2x+1)][(2x+1)/(x+y)]+[(x-y)/(2x+1)][y(2x+1)/(x+y)(x-y)]+y/(x+y)
=(x-y)/(x+y)+y/(x+y)+y/(x+y)
=(x-y+y+y)/(x+y)
=(x+y)/(x+y)
=1
(2x-3)/(2x-4)=1-4/(x^2-4x+4)
(2x-3)/2(x-2)=1-4/(x-2)^2
乘以2(x-2)^2
(2x-3)(x-2)=2(x-2)^2-8
2x^2-7x+6=2x^2-8x+8-8
x=-6
经检验,x=-6时方程的解