Question

TimGTNN của bt

F(x)=\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

Answer 1

phải là GTln chớ bạn

Answer 2

ĐK: \(x\ge1\)

F(x) = \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)}=|\sqrt{x-1}-1|+|\sqrt{x-1}+1|=|1-\sqrt{x-1}|+|\sqrt{x-1}+1|\) \(\le|1-\sqrt{x-1}+\sqrt{x-1}+1|=2\)

Dấu ''='' xảy ra khi \(\left(1-\sqrt{x-1}\right)\left(1+\sqrt{x-1}\right)\ge0\Leftrightarrow1-\sqrt{x-1}\ge0\Leftrightarrow\sqrt{x-1}\le1\Leftrightarrow x\le2\)

Vậy max của F(x) là 2 khi 1 \(\le\) x \(\le2\)