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

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

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

+)Xét \(x< -1\)\(\Rightarrow\begin{cases}x+1< 0\Rightarrow\left|x+1\right|=-\left(x+1\right)=-x-1\\x-1< 0\Rightarrow\left|x-1\right|=-\left(x-1\right)=-x+1\end{cases}\)

\(\Rightarrow y=\left(-x-1\right)-\left(-x+1\right)=2\)

+)Xét \(-1\le x< 1\)\(\Rightarrow\begin{cases}x\ge-1\Rightarrow x+1\ge0\Rightarrow\left|x+1\right|=x+1\\x< 1\Rightarrow x-1< 0\Rightarrow\left|x-1\right|=-\left(x-1\right)=-x+1\end{cases}\)

\(\Rightarrow y=\left(-x+1\right)-\left(x+1\right)=-2x\)

+)Xét \(x\ge1\)\(\Rightarrow\begin{cases}x-1\ge0\Rightarrow\left|x-1\right|=x-1\\x+1\ge0\Rightarrow\left|x+1\right|=x+1\end{cases}\)

\(\Rightarrow y=\left(x-1\right)-\left(x+1\right)=-2\)

Ta thấy:

câu a) rút x theo y thế vào A rồi áp dụng HĐT

b)rút xy thế vào B 

c)HĐT

d)rút x theo y thé vào C

rồi dùng BĐT cô-si

e)BĐT chưa dấu giá trị tuyệt đối

a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)

\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)

b. \(0\le\sqrt{4-x^2}\le2\)

\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)

vậy \(GTNN=\frac{\sqrt{46}}{4}\)

ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)

\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)

Ta có: 

\(C=\sqrt{-x^2+6x}\) 

Mà: \(\sqrt{-x^2+6x}\ge0\) 

Dấu "=" xảy ra khi:

\(\sqrt{-x^2+6x}=0\)

\(\Leftrightarrow\sqrt{-x\left(x-6\right)}=0\)

\(\Leftrightarrow-x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: \(C_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

\(D=\sqrt{6x-2x^2}\)

Mà: \(\sqrt{6x-2x^2}\ge0\)

Dấu "=" xảy ra khi:

\(\sqrt{6x-2x^2}=0\)

\(\Leftrightarrow\sqrt{2x\left(3-x\right)}=0\)

\(\Leftrightarrow2x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy: \(D_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(C=\sqrt{-x^2+6x}=\sqrt{9-\left(x^2-6x+9\right)}=\sqrt{9-\left(x-3\right)^2}\le\sqrt{9}=3\)

Dấu "=" xảy ra khi \(x=3\)

Vậy \(maxC=3\)

\(D=\sqrt{6x-2x^2}=\dfrac{1}{\sqrt{2}}\sqrt{12x-4x^2}=\dfrac{1}{\sqrt{2}}\sqrt{9-\left(4x^2-12x+9\right)}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{9-\left(2x-3\right)^2}\le\dfrac{1}{\sqrt{2}}.\sqrt{9}\)\(=\dfrac{3\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=\dfrac{3}{2}\)

Vậy \(maxD=\dfrac{3\sqrt{2}}{2}\)

ĐKXĐ: \(\dfrac{3}{2}\le x\le3\)

\(A=\sqrt{2x-3}+\sqrt{6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\)

\(A\ge\sqrt{2x-3+6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\ge\sqrt{3}\)

\(A_{min}=\sqrt{3}\) khi \(3-x=0\Rightarrow x=3\)

\(A=1.\sqrt{2x-3}+\sqrt{2}.\sqrt{6-2x}\le\sqrt{\left(1+2\right)\left(2x-3+6-2x\right)}=3\)

\(A_{max}=3\) khi \(2x-3=\dfrac{6-2x}{2}\Rightarrow x=2\)