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+b的最小值是-3(当a=-2,b=-1时取到).解法2:三角换元法a^2+2b^2=6→(a^2)/6+(b^2)/3=1,设a=(根6)cosx,b=(根3)sinx,这里x∈R.a+b=(根3)sinx+(根6)cosx=根号下[(根3)...
|