Question

（x-y）分之一-（x+y）分之一+（x²-y²）分之一（计算）

急!还有2（x-3）-（x-1）²-（x+3）（x-3）=0这道题

Answer 1

1/(x-y)-1/(x+y)+1/(x²-y²)
=x+y/(x²-y²)-(x-y)/(x²-y²)+1/(x²-y²)
=(x+y-x+y+1)/(x²-y²)
=(2y+1)/(x²-y²)

2（x-3）-（x-1）²-（x+3）（x-3）=0
(x-3) (2(x-1）²)-(x+3))=0
得 (x-3) =0 或 2(x-1）²)-(x+3)=0
检验 x=3不符要求
所以2(x-1）²-(x+3)=0
解得 x=1±√2

Answer 2

1/(x-y)-1/(x+y)+1/(x²-y²)
=(x+y-x+y+1)/(x-y)(x+y)
=(2y+1)/(x²-y²)
2（x-3）-（x-1）²-（x+3）（x-3）=0
同时乘以-1得 （x-1）² -2(x-3)+(x-3)(x+3)=0
提供因式： （x-1）² +（x-3）【（x+3）-2】=0
（x-1）² +(x-3)(x+1)=0
展开 x^2-2x+1+x^2-2x-3=0
2x^2-4x-2=0
同时除以2得 x^2-2x-1=0
用公式法求解 x=1±√2
希望你明白

Answer 3

1/(x-y)-1/(x+y)+1/(x²-y²)
=(x+y-x+y+1)/(x-y)(x+y)
=(2y+1)/(x²-y²)

Answer 4

1/(x-y)-1/(x+y)+1/(x²-y²)
=(x+y-x+y+1)/(x-y)(x+y)
=(2y+1)/(x²-y²)