B=x-4\(\sqrt{x}\)+10 (x≥0)
B=x-2.\(\sqrt{x}\).2+4+6
B=(\(\sqrt{x}\)-2)\(^2\)+6
Ta có \(\left(\sqrt{x}-2\right)^2\)≥0 với mọi x tm ĐKXĐ
<=> \(\left(\sqrt{x}-2\right)^2\)+6 ≥6
Dấu = xảy ra <=> \(\left(\sqrt{x}-2\right)^2=0\\ < =>\sqrt{x}-2=0\\ < =>\sqrt{x}=2\\ < =>x=4\left(tm\right)\)
Vậy GTNN B=6 khi x=4