题目

1.计算：1/(m^2-m)+(m-2)/(2m^2-2)
2.用2种不同的运算顺序计算：（x/(x-2)-x/(x+2))*2-x/x（建议一种先算括号里面的,一种用乘法分配率）

提问时间：2020-10-17

答案

1 =2/(2m^2-2)+(m-2)/(2m^2-2)=m/(2m^2-2)
2 ={x*[(x+2)/(x+2)(x-2)+(x-2)/(x+2)(x-2)]}^2-1
=[2x^2/(x+2)(x-2)]^2-1
=[4x^4/(x^2-4)^2]-1
=(3x^4+8x^2-16)/(x^4-8x^2+16)
or
=[x^2/(x-2)^2]-2*x^2/(x+2)(x-2)+[x^2/(x+2)^2]-1
=[x^2(x+2)^2-2x^2(x+2)(x-2)+x^2(x-2)^2]/(x+2)^2(x-2)^2 -1
=[4x^4/(x^2-4)^2]-1
=(3x^4+8x^2-16)/(x^4-8x^2+16)