a) Vì (x-1)2 ≥0, ∀x
=> ( x -1)2 +7 ≥ 7, ∀x
=> \(\sqrt{\left(x-1\right)^2+7}\) ≥\(\sqrt{7}\)
Dấu "=" xảy ra <=> (x - 1)2 = 0
=> x = 1
Vậy min = \(\sqrt{7}\) <=> x=1
b) Đặt A = \(\sqrt{x^2-6x+46}\)
A = \(\sqrt{x^2-6x+9+37}\)
= \(\sqrt{\left(x-3\right)^2+37}\)
Vì (x - 3)2 ≥ 0, ∀x
=> (x -3)2 + 37 ≥ 37, ∀x
=> \(\sqrt{\left(x-3\right)^2+37}\) ≥ \(\sqrt{37}\)
minA =\(\sqrt{37}\) <=> x -3 =0
=> x = 3
Vậy minA = \(\sqrt{37}\) <=> x = 3