1 quy đồng lên ra được

2 \(A=\dfrac{1}{x-2\sqrt{x-5}+3}\le\dfrac{1}{5-2.0+3}=\dfrac{1}{8}\)

dấu"=" xảy ra<=>x=5

ĐKXĐ: x>=4

\(A=\dfrac{1}{x-4\sqrt{x-4}+3}\)

\(=\dfrac{1}{x-4-4\sqrt{x-4}+4+3}\)

\(=\dfrac{1}{\left(\sqrt{x-4}-2\right)^2+3}\)

\(\left(\sqrt{x-4}-2\right)^2+3>=3\)

=>\(A=\dfrac{1}{\left(\sqrt{x-4}-2\right)^2+3}< =\dfrac{1}{3}\)

Dấu = xảy ra khi \(\sqrt{x-4}-2=0\)

=>x-4=4

=>x=8

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

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

 -Nêú \(x\ge1\)thì \(\sqrt{\left(x+1\right)^2}=x+1\)và\(\sqrt{\left(x-1\right)^2}=x-1\)

Ta có:\(A=x+1+x-1=2x\ge2\)

Dấu "=" xảy ra khi x=1

-Nếu\(1>x\ge-1\)thì \(\sqrt{\left(x+1\right)^2}=x+1\)và\(\sqrt{\left(x-1\right)^2}=1-x\)

Ta có:\(A=x+1+1-x=2\)

-Nếu x<-1 thì \(\sqrt{\left(x+1\right)^2}=-x-1\)và\(\sqrt{\left(x-1\right)^2}=1-x\)

Ta có:\(A=-x-1+1-x=-2x\ge2\)

Dấu "=" xảy ra khi x=-1

Vậy GTNN của A là 2 tại x=1 hoặc x=-1

a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)

b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)

\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)

Ta có  x – 2√x + 3 = (√x – 1)² + 2.  Mà (√x – 1)² ≥ 0 với mọi x ≥ 0 ⇒ (√x – 1)² + 2 ≥ 2 với mọi x ≥ 0

⇒ \(A=\frac{1}{\left(\sqrt{X}-1\right)^2+2}\le\frac{1}{2}\)

Vậy GTLN của A = 1/2  ⇔ √x = 1 ⇔ x =1

ủNG HỘ MK NHAK

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

Vậy GTNN là 1/4 khi \(x+\frac{\sqrt{3}}{2}=0\Rightarrow x=-\frac{\sqrt{3}}{2}\)

\(A=\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-1\right)^2}\)

\(=\left|x+1\right|+\left|x-1\right|=\left|x+1\right|+\left|1-x\right|\)

Áp dụng bđt \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:

\(A\ge\left|x+1+1-x\right|=2\)

Vậy GTNN của A là 2 khi \(-1\le x\le1\)

Ta có 

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

\(\Rightarrow A=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-1\right)^2}\)

\(\Rightarrow A=\left|x+1\right|+\left|x-1\right|\)

\(\Rightarrow A=\left|x+1\right|+\left|1-x\right|\)

Vì \(\begin{cases}\left|x+1\right|\ge x+1\\\left|1-x\right|\ge1-x\end{cases}\)\(\Rightarrow\left|x+1\right|+\left|1-x\right|\ge x+1+1-x\)

\(\Rightarrow\left|x+1\right|+\left|1-x\right|\ge2\)

Dấu " = " xảy ra khi \(\begin{cases}x+1\ge0\\1-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-1\\x\le1\end{cases}\)

Vậy MIN_A=2 khi \(-1\le x\le1\)