作业帮1/x(x-1)² 1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]1/x(x-1)²1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]我不理解为何拆成上面的形式,我自己拆就会拆成A／x+B／（x-1）*2或是A／x+B／x+1+C／x-1.自学有疑惑,
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 05:04:10
![1/x(x-1)² 1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]1/x(x-1)²1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]我不理解为何拆成上面的形式,我自己拆就会拆成A／x+B／（x-1）*2或是A／x+B／x+1+C／x-1.自学有疑惑,](/uploads/image/z/2555722-10-2.jpg?t=1%2Fx%28x-1%29%26%23178%3B+1%2Fx%28x-1%29%26%23178%3B%3D%5BA%2Fx%5D%2B%5BB%2F%28x-1%29%26%23178%3B%5D%2B%5BC%2F%28x-1%29%5D1%2Fx%28x-1%29%26%23178%3B1%2Fx%28x-1%29%26%23178%3B%3D%5BA%2Fx%5D%2B%5BB%2F%28x-1%29%26%23178%3B%5D%2B%5BC%2F%28x-1%29%5D%E6%88%91%E4%B8%8D%E7%90%86%E8%A7%A3%E4%B8%BA%E4%BD%95%E6%8B%86%E6%88%90%E4%B8%8A%E9%9D%A2%E7%9A%84%E5%BD%A2%E5%BC%8F%2C%E6%88%91%E8%87%AA%E5%B7%B1%E6%8B%86%E5%B0%B1%E4%BC%9A%E6%8B%86%E6%88%90A%EF%BC%8Fx%2BB%EF%BC%8F%EF%BC%88x-1%EF%BC%89%2A2%E6%88%96%E6%98%AFA%EF%BC%8Fx%2BB%EF%BC%8Fx%2B1%2BC%EF%BC%8Fx-1.%E8%87%AA%E5%AD%A6%E6%9C%89%E7%96%91%E6%83%91%2C)
1/x(x-1)² 1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]1/x(x-1)²1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]我不理解为何拆成上面的形式,我自己拆就会拆成A／x+B／（x-1）*2或是A／x+B／x+1+C／x-1.自学有疑惑,
1/x(x-1)² 1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]
1/x(x-1)²
1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]
我不理解为何拆成上面的形式,我自己拆就会拆成A／x+B／（x-1）*2或是A／x+B／x+1+C／x-1.自学有疑惑,求教这种有理函数拆分的原理,这题是怎么拆的.
1/x(x-1)² 1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]1/x(x-1)²1/x(x-1)²=[A/x]+[B/(x-1)²]+[C/(x-1)]我不理解为何拆成上面的形式,我自己拆就会拆成A／x+B／（x-1）*2或是A／x+B／x+1+C／x-1.自学有疑惑,
将后面的代数式化为(表达式)/x(x-1)^2 ==>表达式中二次项和一次项为0,常数为1,
最后得出结果为A=1,B=1,C=-1