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

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