分式4x-5\2x^2+x-6是有两个分式m\x+2与n\2x-3相加得到的,求m,n的值

Question

分式4x-5\2x^2+x-6是有两个分式m\x+2与n\2x-3相加得到的,求m,n的值

Answer 1

因为m/(x+2)+n/(2x-3)
=m(2x-3)/(x+2)(2x-3)+n(x+2)/(x+2)(2x-3)
=(2mx-3m)/(x+2)(2x-3)+(nx+2n)/(x+2)(2x-3)
=(2mx-3m+nx+2n)/(x+2)(2x-3)
=[(2m+n)x+2n-3m]/(x+2)(2x-3)
所以2m+n=4,2n-3m=-5,
解得m=13/7,n=2/7