a. 

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x(x-2)=0\end{matrix}\right.\Leftrightarrow x=2\)

b. ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$

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

$\Leftrightarrow |\sqrt{x-1}+1|=2$

$\Leftrightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow \sqrt{x-1}=1$

$\Leftrightarrow x=2$ (tm)

c. 

PT \(\Leftrightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x+1=4x^2-4x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=2x(x-1)=0\end{matrix}\right.\Leftrightarrow x=1\) (tm)

d.

ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$

$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$

$\Leftrightarrow \sqrt{x-4}+2=2$

$\Leftrightarrow \sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

a: Ta có: \(\sqrt{x^2-2x+4}=2x-2\)

\(\Leftrightarrow x^2-2x+4=4x^2-8x+4\)

\(\Leftrightarrow-3x^2+6x=0\)

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

\(\Leftrightarrow x=2\)

b: Ta có: \(\sqrt{x+2\sqrt{x-1}}=2\)

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

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

\(\Leftrightarrow x-1=1\)

hay x=2

c: Ta có: \(\sqrt{2x^2-2x+1}=2x-1\)

\(\Leftrightarrow2x^2-2x+1=4x^2-4x+1\)

\(\Leftrightarrow-2x^2+2x=0\)

\(\Leftrightarrow-2x\left(x-1\right)=0\)

hay x=1

Ta có: \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

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

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

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

\(=2\sqrt{2}\)

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)

a) ĐKXĐ: \(3\le x\le10\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-4\\x\ne4\end{matrix}\right.\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ne4\end{matrix}\right.\)

d) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

e) ĐKXĐ: \(x\in R\)

1) Ta có: \(\left|x^2-4x-5\right|=x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x-1\left(\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\right)\\-x^2+4x+5=x-1\left(-1< x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5-x+1=0\\-x^2+4x+5-x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=0\\-x^2+3x+6=0\end{matrix}\right.\Leftrightarrow x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{41}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{\sqrt{41}}{2}\\x-\dfrac{5}{2}=-\dfrac{\sqrt{41}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}+5}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}+5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{41}+5}{2}\right\}\)