Đáp án:
$\begin{array}{l}
b)Dkxd:x \ge \dfrac{5}{2}\\
x - \sqrt {2x - 5} = 4\\
\Leftrightarrow x - 4 = \sqrt {2x - 5} \left( {dkxd:x \ge 4} \right)\\
\Leftrightarrow {x^2} - 8x + 16 = 2x - 5\\
\Leftrightarrow {x^2} - 10x + 21 = 0\\
\Leftrightarrow {x^2} - 3x - 7x + 21 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
x = 7\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = 7\\
b)\left( {x - 3} \right).\sqrt {{x^2} + 4} = {x^2} - 9\\
\Leftrightarrow \left( {x - 3} \right)\left( {\sqrt {{x^2} + 4} } \right) = \left( {x - 3} \right).\left( {x + 3} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
\sqrt {{x^2} + 4} = x + 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
{x^2} + 4 = {x^2} + 6x + 9\left( {x \ge - 3} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
6x = - 5\left( {x \ge - 3} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{{ - 5}}{6}\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = 3;x = - \dfrac{5}{6}\\
f)Dkxd:x \le \dfrac{3}{2}\\
x - \sqrt {3 - 2x} \ge 0\\
\Leftrightarrow x \ge \sqrt {3 - 2x} \left( {x \ge 0} \right)\\
\Leftrightarrow {x^2} \ge 3 - 2x\\
\Leftrightarrow {x^2} + 2x - 3 \ge 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 1} \right) \ge 0\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le - 3
\end{array} \right.\\
Khi:0 \le x \le \dfrac{3}{2}\\
\Leftrightarrow 1 \le x \le \dfrac{3}{2}\\
Vậy\,1 \le x \le \dfrac{3}{2}\\
h)\sqrt { - 3x + 2} + 1 < x\\
\Leftrightarrow \sqrt { - 3x + 2} < x - 1\\
Dkxd:\left\{ \begin{array}{l}
- 3x + 2 \ge 0\\
x - 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{2}{3}\\
x \ge 1
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
j)Dkxd:\left\{ \begin{array}{l}
x + 1 \ge 0\\
{x^2} - 4x + 3 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
\left( {x - 1} \right)\left( {x - 3} \right) \ge 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 1 \le x \le 1\\
x \ge 3
\end{array} \right.\\
\sqrt {{x^2} - 4x + 3} < x + 1\\
\Leftrightarrow {x^2} - 4x + 3 < {x^2} + 2x + 1\\
\Leftrightarrow 6x > 2\\
\Leftrightarrow x > \dfrac{1}{3}\\
Vậy\,\dfrac{1}{3} < x \le 1\,hoặc\,x \ge 3
\end{array}$
