Đáp án:

\(\begin{array}{l}
a)\,\,\,\,S = \left\{ {1 + \sqrt 6 ; - 3 - 2\sqrt 3 } \right\}.\\
b)\,\,\,\,S = \left\{ {0;1} \right\}.\\
c)\,\,S = \left\{ {1; - \frac{7}{2}} \right\}.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\,\,\sqrt {16{x^2} + 8x + 1}  = {x^2} + 2x - 4\\
 \Leftrightarrow \sqrt {{{\left( {4x + 1} \right)}^2}}  = {x^2} + 2x - 4\\
 \Leftrightarrow \left| {4x + 1} \right| = {x^2} + 2x - 4\\
TH1:\,\,4x + 1 \ge 0 \Leftrightarrow x \ge  - \frac{1}{4}\\
pt \Rightarrow 4x + 1 = {x^2} + 2x - 4\\
 \Leftrightarrow {x^2} - 2x - 5 = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 1 + \sqrt 6 \,\left( {tm} \right)\\
x = 1 - \sqrt 6 \,\,\left( {ktm} \right)
\end{array} \right.\\
TH2:\,\,4x + 1 < 0 \Leftrightarrow x <  - \frac{1}{4}\\
pt \Rightarrow  - 4x - 1 = {x^2} + 2x - 4\\
 \Leftrightarrow {x^2} + 6x - 3 = 0 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3 + 2\sqrt 3 \,\,\left( {ktm} \right)\\
x =  - 3 - 2\sqrt 3 \,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,\,S = \left\{ {1 + \sqrt 6 ; - 3 - 2\sqrt 3 } \right\}.\\
b)\,\,1 + \frac{2}{3}\sqrt {x - {x^2}}  = \sqrt x  + \sqrt {1 - x} \\
DK:\,\,\left\{ \begin{array}{l}
x - {x^2} \ge 0\\
x \ge 0\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow 0 \le x \le 1\\
Dat\,\,t = \sqrt x  + \sqrt {1 - x}  \Rightarrow {t^2} = x + 1 - x + 2\sqrt {x\left( {1 - x} \right)} \\
 \Rightarrow {t^2} = 1 + 2\sqrt {x - {x^2}}  \Rightarrow 2\sqrt {x - {x^2}}  = {t^2} - 1\\
PT:\,\,1 + \frac{1}{3}\left( {{t^2} - 1} \right) = t \Leftrightarrow {t^2} - 1 + 3 = 3t\\
 \Leftrightarrow {t^2} - 3t + 2 = 0 \Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = 2
\end{array} \right.\\
 + )\,\,Voi\,\,t = 1 \Rightarrow \sqrt x  + \sqrt {1 - x}  = 1\\
 \Leftrightarrow 1 + 2\sqrt {x - {x^2}}  = 1 \Leftrightarrow \sqrt {x - {x^2}}  = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\,\,\left( {tm} \right)\\
 + )\,\,Voi\,\,t = 2 \Rightarrow \sqrt x  + \sqrt {1 - x}  = 2\\
 \Leftrightarrow 1 + 2\sqrt {x - {x^2}}  = 4 \Leftrightarrow \sqrt {x - {x^2}}  = \frac{3}{2}\\
 \Leftrightarrow {x^2} - x + \frac{9}{4} = 0\,\,\left( {Vo\,\,nghiem} \right)\\
Vay\,\,S = \left\{ {0;1} \right\}.\\
c)\,\,\sqrt {2{x^2} + 5x + 2}  - 2\sqrt {2{x^2} + 5x - 6}  = 1\\
Dat\,\,\sqrt {2{x^2} + 5x - 6}  = u\,\,\left( {u \ge 0} \right)\\
 \Rightarrow {u^2} = 2{x^2} + 5x - 6 \Rightarrow {u^2} + 8 = 2{x^2} + 5x + 2\\
PT:\,\,\sqrt {{u^2} + 8}  - 2u = 1\\
 \Leftrightarrow \sqrt {{u^2} + 8}  = 2u + 1\\
 \Leftrightarrow {u^2} + 8 = {\left( {2u + 1} \right)^2}\,\,\left( {Do\,\,u \ge 0 \Rightarrow 2u + 1 > 0} \right)\\
 \Leftrightarrow {u^2} + 8 = 4{u^2} + 4u + 1\\
 \Leftrightarrow 3{u^2} + 4u - 7 = 0 \Leftrightarrow \left[ \begin{array}{l}
u = 1\,\,\left( {tm} \right)\\
u =  - \frac{7}{3}\,\,\left( {ktm} \right)
\end{array} \right.\\
u = 1 \Rightarrow \sqrt {2{x^2} + 5x - 6}  = 1\\
 \Leftrightarrow 2{x^2} + 5x - 6 = 1 \Leftrightarrow 2{x^2} + 5x - 7 = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \frac{7}{2}
\end{array} \right.\\
Vay\,\,S = \left\{ {1; - \frac{7}{2}} \right\}.
\end{array}\)