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}\)