Question

设a、b属于R,a^2+2b^2=6,则a+b的最小值是____

Answer 1

解法1：判别式法.设a+b＝t,则a＝t-b.[1] 代入条件得：(t-b)^2+2b^2=6,3b^2-2tb+(t^2-6)=0.[2] ∵b是实数,∴判别式Δ≥0,即4t^2-12(t^2-6)≥0,化简得：t^2≤9,∴-3≤t≤3.当t＝-3时,由[2]得b=-1,代入[1]得a=-2.所以a+...