a)\(đkx\ge1,x\ne-1\)

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

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

\(\Leftrightarrow x-1=4x-4\)

\(\Leftrightarrow x=1\)(nhận)

Vậy S=\(\left\{1\right\}\)

c)đk\(25x^2-10x+1=\) \(\left(5x-1\right)^2\ge0\Leftrightarrow x\ge\dfrac{1}{5}\)

\(\sqrt{25x^2-10x+1}+2x=1\)

\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}+2x=1\)

\(\Leftrightarrow5x-1+2x=1\)

\(\Leftrightarrow x=\dfrac{2}{7}\)(nhận)

Vậy S=\(\left\{\dfrac{2}{7}\right\}\)

c: Ta có: \(\sqrt{25x^2-10x+1}+2x=1\)

\(\Leftrightarrow\left|5x-1\right|=1-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=1-2x\left(x\ge\dfrac{1}{5}\right)\\5x-1=2x-1\left(x< \dfrac{1}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)

\(1,=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]\\ =2\left(x+y+1\right)\left(x-y+1\right)\\ 5,=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

2) \(=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\)

4) \(=2\left[\left(x^2+2x+1\right)-y^2\right]=2\left[\left(x+1\right)^2-y^2\right]\)

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

5) \(=16-\left(x^2-2xy+y^2\right)=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)