已知４比Ｘ平方减１的差等于Ａ比Ｘ减一的差加Ｂ比Ｘ加１的和是恒等式,则Ａ＝?Ｂ＝?

首先,将右边的式子进行通分,得：右式=A(x+1)/(x-1)+B(x-1)/(x+1)=[A(x+1)+B(x-1)]/(x^2-1)=[(A+B)x+(A-B)]/(x^2-1) 这时,左式=4/(x^2-1),即可得：方程组：A+B=0；A-B=4,解方程组可得：A=2,B=-2