Question

1, tìm min , max

a, \(A=3\sqrt{x-1}+7\)

b,\(\frac{4}{\sqrt{x}+3}\)

c,\(C=\frac{3\sqrt{x}+8}{\sqrt{x}+3}\)

d,\(D=x-3\sqrt{x}+2\)

e, \(E=\frac{4}{x-2\sqrt{x}+3}\)

ai nhanh tick nhé <3 giúp mình nèo

Answer 1

Lời giải :

a) \(A=3\sqrt{x-1}+7\ge7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=1\)

b) \(B=\frac{4}{\sqrt{x}+3}\le\frac{4}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

c) \(C=\frac{3\sqrt{x}+8}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-1}{\sqrt{x}+3}=3-\frac{1}{\sqrt{x}+3}\)

Có \(\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\forall x\)

\(\Leftrightarrow-\frac{1}{\sqrt{x}+3}\ge\frac{-1}{3}\)

\(\Leftrightarrow3-\frac{1}{\sqrt{x}+3}\ge3-\frac{1}{3}=\frac{8}{3}\)

\(\Leftrightarrow C\ge\frac{8}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

d) \(D=x-3\sqrt{x}+2\)

\(D=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{3}{2}+\frac{9}{4}-\frac{1}{4}\)

\(D=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)

e) \(E=\frac{4}{x-2\sqrt{x}+3}=\frac{4}{\left(\sqrt{x}-1\right)^2+2}\le\frac{4}{2}=2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

Answer 2

a) Vì \(3\sqrt{x-1}\ge0\forall x\ge1\) 

 \(\Rightarrow3\sqrt{x-1}+7\ge7\forall x\ge1\) 

Dấu "=" xảy ra <=>\(3\sqrt{x-1}=0\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) 

Vậy A_min =7 tại x=1