已知：a+x^2=2000,b+x^2=2001,c+x^2=2002,且abc=2求a/bc+c/ab+b/ac-1/a-1/b-1/c的值

=(a²＋b²＋c²)/abc－(ab＋bc＋ca)/abc=[a²＋b²＋c²－ab－bc－ca]/2=(1/2)[(a²－ab)＋(b²－bc)＋(c²－ca)]=(1/2)[a(a－b)＋b(b－c)＋c(c－a)]=(1/2)[－a－b＋2c]=(1/2)[(c－a)＋(c－b)]=3/2.如何得到的(1/2)[－a－b＋2c]