函数f(x)=ax+1/x+2在区间(-2,+无穷)上单调递增,则实数a的取值范围是解得f(x)=（ax+1)/(x+2)=[a(x+2)+1-2a]/(x+2)=a+(1-2a)/(x+2)a>1/2.就是想问一下为什么[a(x+2)+1-2a]/(x+2)会等于a+(1-2a)/(x+2),它是怎么化简的?

Question

函数f(x)=ax+1/x+2在区间(-2,+无穷)上单调递增,则实数a的取值范围是
解得f(x)=（ax+1)/(x+2)=[a(x+2)+1-2a]/(x+2) =a+(1-2a)/(x+2) a>1/2 .就是想问一下为什么[a(x+2)+1-2a]/(x+2)会等于a+(1-2a)/(x+2),它是怎么化简的?

Answer 1

[a(x+2)+1-2a]/(x+2) =[a(x+2)+（1-2a）]/(x+2) =a(x+2)/(x+2)+(1-2a)/(x+2) = a + (1-2a)/(x+2)
这里a是单独分开的的,不在分子那里