Đáp án:
14) x=2
Giải thích các bước giải:
\(\begin{array}{l}
13)DK:x\left[ \begin{array}{l}
x \ge \sqrt 5 \\
x \le - \sqrt 5
\end{array} \right.\\
\sqrt {{x^2} - 5} = 2 - x\\
\to {x^2} - 5 = 4 - 4x + {x^2}\left( {DK:2 \ge x} \right)\\
\to 4x = 9\\
\to x = \dfrac{9}{4}\left( {KTM} \right)\\
\to x \in \emptyset \\
14)\sqrt {x - 1} = 2x - 3\\
\to x - 1 = 4{x^2} - 12x + 9\left( {DK:x \ge \dfrac{3}{2}} \right)\\
\to 4{x^2} - 13x + 10 = 0\\
\to \left[ \begin{array}{l}
x = 2\left( {TM} \right)\\
x = \dfrac{5}{4}\left( l \right)
\end{array} \right.\\
15)DK:x \ne 0\\
TH1:\left| {2x - \dfrac{1}{x}} \right| = 2x - \dfrac{1}{x}\left( {DK:2x - \dfrac{1}{x} \ge 0} \right)\\
Pt \to {\left( {2x - \dfrac{1}{x}} \right)^2} + \left( {2x - \dfrac{1}{x}} \right) - 6 = 0\\
\to \left[ \begin{array}{l}
2x - \dfrac{1}{x} = 2\\
2x - \dfrac{1}{x} = - 3\left( l \right)
\end{array} \right.\\
\to 2{x^2} - 2x - 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt 3 }}{2}\\
x = \dfrac{{1 - \sqrt 3 }}{2}
\end{array} \right.\\
TH2:\left| {2x - \dfrac{1}{x}} \right| = - \left( {2x - \dfrac{1}{x}} \right)\left( {DK:2x - \dfrac{1}{x} < 0} \right)\\
Pt \to {\left( {2x - \dfrac{1}{x}} \right)^2} - \left( {2x - \dfrac{1}{x}} \right) - 6 = 0\\
\to \left[ \begin{array}{l}
2x - \dfrac{1}{x} = 3\left( l \right)\\
2x - \dfrac{1}{x} = - 2
\end{array} \right.\\
\to 2{x^2} + 2x - 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt 3 }}{2}\\
x = \dfrac{{1 - \sqrt 3 }}{2}
\end{array} \right.\\
KL:\left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt 3 }}{2}\\
x = \dfrac{{1 - \sqrt 3 }}{2}
\end{array} \right.\\
16)DK:x \ge - 3\\
Pt \to \left[ \begin{array}{l}
{x^2} + 5x + 4 = 0\\
x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x + 1} \right)\left( {x + 4} \right) = 0\\
x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( {TM} \right)\\
x = - 4\left( l \right)\\
x = - 3\left( {TM} \right)
\end{array} \right.
\end{array}\)
