Đáp án:
a) \(S = \left\{ { - 1;3;7} \right\}\).
b) \(S = \left\{ 1 \right\}\).
c) \(S = \left\{ 2 \right\}\).
Giải thích các bước giải:
a) \(\left| {{x^2} - 4x - 5} \right| = 2\left| {x + 1} \right|\).
\( \Rightarrow \left[ \begin{array}{l}{x^2} - 4x - 5 = 2x + 2\\{x^2} - 4x - 5 = - 2x - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x^2} - 6x - 7 = 0\\{x^2} - 2x - 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 7\\x = - 1\\x = 3\end{array} \right.\)
Vậy \(S = \left\{ { - 1;3;7} \right\}\).
b) \(\sqrt {{x^2} - 2x + 5} = {x^2} - 2x + 3\)
Đặt
\(\begin{array}{l}t = \sqrt {{x^2} - 2x + 5} = \sqrt {{{\left( {x - 1} \right)}^2} + 4} \ge 2\\ \Rightarrow {x^2} - 2x + 5 = {t^2} \Rightarrow {x^2} - 2x + 3 = {t^2} - 2\end{array}\)
Phương trình trở thành \(t = {t^2} - 2 \Leftrightarrow {t^2} - t - 2 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 2\,\,\left( {tm} \right)\\t = - 1\,\,\left( {ktm} \right)\end{array} \right.\).
\(\begin{array}{l}t = 2 \Rightarrow \sqrt {{x^2} - 2x + 5} = 2 \Leftrightarrow {x^2} - 2x + 5 = 4\\ \Leftrightarrow {x^2} - 2x + 1 = 0 \Leftrightarrow {\left( {x - 1} \right)^2} = 0 \Leftrightarrow x = 1\end{array}\)
Vậy \(S = \left\{ 1 \right\}\).
c) \(\sqrt {x - 2} \left( {{x^2} - 3x + 2} \right) = 0\) (ĐK: \(x \ge 2\))
\( \Leftrightarrow \left[ \begin{array}{l}x - 2 = 0\\{x^2} - 3x + 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 2\,\,\left( {tm} \right)\\x = 1\,\,\left( {ktm} \right)\end{array} \right.\).
Vậy \(S = \left\{ 2 \right\}\).
