要y=根号(3-x)+根号(10-2x)有意义须,3-x≥0且10-2x≥0,即x≤3
而当x≤3时,3-x≥0,10-2x≥4,根号(3-x)+根号(10-2x）≥0+√4=2
故所求最小值为2
补充：若y=根号(x-3)+根号(10-2x),
由x-3≥0且10-2x≥0有
3≤x≤5
-1≤x-4≤1
令x-4=sinθ,-π/2≤θ≤π/2,则x=4+sinθ,-π/4≤θ/2≤π/4,
1/√2≤cos(θ/2)≤1
-1/√2≤sin(θ/2)≤1/√2
cos(θ/2)+sin(θ/2)≥0
cos(θ/2)-sin(θ/2)≥0
y=根号(1+sinθ)+根号(2-2sinθ)
=根号[cos²(θ/2)+sin²(θ/2)+2cos(θ/2)sin(θ/2)]+根号{2[cos²(θ/2)+sin²(θ/2)-2cos(θ/2)sin(θ/2)]}
=cos(θ/2)+sin(θ/2)+√2cos(θ/2)-√2sin(θ/2)
=(√2+1)cos(θ/2)-(√2-1)sin(θ/2)
=√6cos(θ/2+α)
α=arccos(1/√3+1/√6)
θ/2=π/4时,θ=π/2,x=5,y=根号(x-3)+根号(10-2x)取最小值√2