Question

[Nguyễn Hoàng Tiến]

Tim Min \(A=\sqrt{x}+\sqrt{2-x}\)

Dau tien ta chung minh BDT \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\)

That vay 2 ve luon duong nen \(\left(\sqrt{A}+\sqrt{B}\right)^2\ge\left(\sqrt{A+B}\right)^2\)

<=> \(A+B+2\sqrt{AB}\ge A+B\)

<=> \(2\sqrt{AB}\ge0\) (dieu nay dung vi A va B luon duong hoac bang 0)

<=> \(AB\ge0\) day la dau bang cua BDT

Ap dung, ta co: \(\sqrt{x}+\sqrt{2-x}\ge\sqrt{x+2-x}=\sqrt{2}\)

Dau bang <=> \(x\left(2-x\right)\ge0\)

*TH1: \(x\ge0;2-x\ge0\Leftrightarrow0\le x\le2\)

*TH2: \(x\le0;2-x\le0\Leftrightarrow0\le x;x\ge2\Leftrightarrow x\in\)rong

Vay \(\sqrt{x}+\sqrt{2-x}\ge\sqrt{2}\Leftrightarrow0\le x\le2\)

Answer 1

[s2 Lắc Lư s2]

khỏi cần

ta có \(A^2=2+2\sqrt{x\left(2-x\right)}\ge2\)

dấu = xảy ra khi x=4

Answer 2

[s2 Lắc Lư s2]

nhanh hơn nhìu nha