Đáp án:
e. \(\left[ \begin{array}{l}
x = 2\\
x = - 7
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 1\\
\sqrt {x - 1} + 2\sqrt {x - 1} - 5\sqrt {x - 1} + 2 = 0\\
\to \left( {1 + 2 - 5} \right)\sqrt {x - 1} + 2 = 0\\
\to 2\sqrt {x - 1} = 2\\
\to \sqrt {x - 1} = 1\\
\to x - 1 = 1\\
\to x = 2\left( {TM} \right)\\
b.DK:x \ge 1\\
\dfrac{1}{2}\sqrt {x - 1} - \dfrac{3}{2}.3\sqrt {x - 1} + 24.\dfrac{{\sqrt {x - 1} }}{8} = - 17\\
\to - \sqrt {x - 1} = - 17\\
\to \sqrt {x - 1} = 17\\
\to x - 1 = 289\\
\to x = 290\\
3)3\sqrt {{x^2} + 2} + 2\sqrt {{x^2} + 2} - 5\sqrt {{x^2} + 2} + 3 = 0\\
\to 0\sqrt {{x^2} + 2} + 3 = 0\left( {vô lý} \right)\\
\to x \in \emptyset \\
4) - {x^2} + 2x + \sqrt {6{x^2} - 12x + 7} = 0\\
\to \sqrt {6{x^2} - 12x + 7} = {x^2} - 2x\left( 1 \right)\\
Đặt:\sqrt {6{x^2} - 12x + 7} = t\left( {t \ge 0} \right)\\
\to 6{x^2} - 12x + 7 = {t^2}\\
\to 6{x^2} - 12x = {t^2} - 7\\
\to {x^2} - 2x = \dfrac{{{t^2} - 7}}{6}\\
Pt\left( 1 \right) \to t = \dfrac{{{t^2} - 7}}{6}\\
\to {t^2} - 6t - 7 = 0\\
\to \left( {t - 7} \right)\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = 7\\
t = - 1\left( l \right)
\end{array} \right.\\
\to \sqrt {6{x^2} - 12x + 7} = 7\\
\to 6{x^2} - 12x + 7 = 49\\
\to 6{x^2} - 12x - 42 = 0\\
\to \left[ \begin{array}{l}
x = 1 + 2\sqrt 2 \\
x = 1 - 2\sqrt 2
\end{array} \right.\\
e.DK:\left[ \begin{array}{l}
x \ge \dfrac{{ - 5 + \sqrt {17} }}{2}\\
x \le \dfrac{{ - 5 - \sqrt {17} }}{2}
\end{array} \right.\\
{x^2} + 5x + 4 - 3\sqrt {{x^2} + 5x + 2} = 6\left( 2 \right)\\
Đặt:\sqrt {{x^2} + 5x + 2} = t\left( {t \ge 0} \right)\\
\to {x^2} + 5x + 2 = {t^2}\\
\to {x^2} + 5x + 4 = {t^2} + 2\\
\left( 2 \right) \to {t^2} + 2 - 3t = 6\\
\to {t^2} - 3t - 4 = 0\\
\to \left( {t - 4} \right)\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = 4\\
t = - 1\left( l \right)
\end{array} \right.\\
\to \sqrt {{x^2} + 5x + 2} = 4\\
\to {x^2} + 5x + 2 = 16\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 7
\end{array} \right.\left( {TM} \right)
\end{array}\)
