\(a,DKXD:x\ge0\)

\(b,A=\sqrt{x-\sqrt{x^2-4x+4}}\)

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

\(=\sqrt{x-\left|x-2\right|}\)

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

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

\(=\sqrt{2}\)

a.\(DKXD:x\ge1\)

b.\(A=\sqrt{x-\sqrt{x^2-4x+4}}=\sqrt{x-\sqrt{\left(x-2\right)^2}}=\sqrt{x-|x-2|}=\orbr{\begin{cases}\sqrt{2}\left(x\ge2\right)\\2x-2\left(1\le x< 2\right)\end{cases}}\)

\(đkxđ\Leftrightarrow x\ge\sqrt{x^2-4x+4}\)\(\Rightarrow x\ge|x-2|\Rightarrow x\ge0\)

\(A=\sqrt{x-\sqrt{x^2-4x+4}}.\)

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

\(=\sqrt{x-|x-2|}=0\)

Nếu \(x\ge2\Rightarrow A=\sqrt{x-\left(x-2\right)}=\sqrt{x-x+2}=\sqrt{2}\)

Nếu \(0\le x< 2\Rightarrow A=\sqrt{x-\left(2-x\right)}=\sqrt{2x-2}\)

\(b,\sqrt{\frac{2x-1}{x+3}}\)

\(Đk:\)\(x+3\ne0\Rightarrow x\ne-3\)

Và \(\frac{2x-1}{x+3}\ge0\)

Khi \(\frac{2x-1}{x+3}=0\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Khi \(\frac{2x-1}{x+3}>0\)\(\Rightarrow\orbr{\begin{cases}2x-1>0;x+3>0\\2x-1< 0;x+3< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2};x>-3\\x< \frac{1}{2};x< -3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}}\)

Vậy căn thức xác định khi \(x\ge\frac{1}{2};x< -3\)

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

Bài 1:

ĐKXĐ: $3-2x\geq 0\Leftrightarrow x\leq \frac{3}{2}$

Bài 2:

a. ĐKXĐ: $x\geq \frac{1}{3}$

PT $\Leftrightarrow 3x-1=2^2=4$

$\Leftrightarrow x=\frac{5}{3}$ (tm)

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{x-2}+2\sqrt{x-2}=6$

$\Leftrightarrow 3\sqrt{x-2}=6$

$\Leftrightarrow \sqrt{x-2}=2$

$\Leftrightarrow x-2=4$

$\Leftrightarrow x=6$ (tm)