解析:依题意

由②得y=≥1,

∵x+10>0,∴x(x+40)≥15(x+10).

∴x2+25x-150≥0.

∴(x+30)(x-5)≥0.

∵x+30>0,∴x-5≥0,即x≥5.

答:第一次至少买5件商品.

备选习题

11若x

解析:(用作差法比较)

(x2+y2)(x-y)-(x2-y2)(x+y)

=(x-y)［(x2+y2)-(x+y)2］=-2xy(x-y).

∵x0,x-y<0.

∴-2xy(x-y)>0.

∴(x2+y2)(x-y)>(x2-y2)(x+y).

12令0

A. B.a C.2ab D.a2+b2

解析:由题意,0

又由2ab≤≤a2+b2,

得a<2ab<

答案:D

13给出函数f(x)=x2,对任意x1,x2∈R+,且x1≠x2,试比较［f(x1)+f(x2)］与f()的大小关系.

解析:∵［f(x1)+f(x2)］-f()

= (x12+x22)-()2

=x12+x22-x12-x22-x1x2