ban nen phan tich x^5+1 thanh:
(x+1)[x^2+(-1-canbachaicua5)/2.x+1][x^2+(-1+canbachaicua5)/2.x+1]tiep denphan tich:
1/(x^5+1)thanh A/(x+1)+ (Bx+C)/[x^2+(-1-canbachaicua5)/2.x+1]+
(Dx+E)/[x^2+(-1+canbachaicua5)/2.x+1]
dung dong nhat thuc tim cac he so A,B,C,D,E roi tiep theo ban thuc hien not cong doan cuoi cung, bye
giup tui bai tich phan nay voi
Posted - 20/03/2003 : 11:15:23 
