作业帮[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)],
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/02 18:35:44
![[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)],](/uploads/image/z/1633951-55-1.jpg?t=%5B%28x%26%23178%3B-2x%2B1%E5%88%86%E4%B9%8B1-x%26%23178%3B%2B2x-%2B1%E5%88%86%E4%B9%8B1%29%5D%C3%B7%28x%2B1%E5%88%86%E4%B9%8B1%29%2B%5B%28x-1%E5%88%86%E4%B9%8B1%29%5D%2C)
[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)],
[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)],
[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)],
[(x²-2x+1分之1-x²+2x-+1分之1)]÷(x+1分之1)+[(x-1分之1)]
=[1/(x-1)²-1/(x+1)²]÷[1/(x+1）+1/(x-1)]
=[1/(x+1）+1/(x-1)][1/(x-1）+1/(x+1)]÷[1/(x+1）+1/(x-1)]
=1/(x-1）-1/(x+1)
=(x+1)/(x+1）(x-1)-(x-1)/(x-1)(x+1)
=(x+1-x+1)/(x-1)(x+1)
=2/(x+1)(x-1)
提示：用平方差公式,不要通分