分式的加减：1/(x-5) - 6/(x^2-25) -(x+1)/(10+2x)
问题描述：
分式的加减：1/(x-5) - 6/(x^2-25) -(x+1)/(10+2x)
答
=2(x+5)/2(x+5)(x-5)-12/2(x+5)(x-5)-(x+1)(x-5)/2(x+5)(x-5)
=[2(x+5)-12-(x+1)(x-5)]/2(x+5)(x-5)
=(2x+10-12-x²+4x+5)/2(x+5)(x-5)
=(-x²+6x+15)/(2x²-50)
答
1/(x-5)-6/(x^2-25)-(x+1)/(10+2x)
=2(x+5)/2(x^2-25)-12/2(x^2-25)-2(x+1)(x-5)/2(x^2-25)
=[2x+10-12-2(x^2-4x-5)]/2(x^2-25)
=(2x-2-2x^2+6x+10)/2(x^2-25)
=(8x-2x^2+8)/2(x^2-25)
=(4x-x^2+4)/(x^2-25)