A = \(\sqrt{\left(x-3\right)-2\sqrt{x-3}+1+2}\)
= \(\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\)\(\ge\)\(\sqrt{0+2}\)=\(\sqrt{2}\)
''='' <=> x = 4
=> Min A = \(\sqrt{2}\)và x = 4
B = |x-2011| + |x-1|
TH1: x \(\le\)1
=> B = 2012 - 2x \(\ge\)2010 ''='' <=> x = 1
TH2: 1\(\le\)x\(\le\)2011
=> B = x - 1 + 2011 - x = 2010 với mọi x t/m đkiện
TH3: x \(\ge\)2011
=> B = 2x - 2012 \(\ge\)2010 ''='' <=> x = 2011
Vậy Min B = 2010 <=> 1\(\le\)x\(\le\)2011