Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
2d,\\
\sqrt {x + 1} - \sqrt {x - 2} = 1\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow \sqrt {x + 1} = \sqrt {x - 2} + 1\\
\Leftrightarrow {\sqrt {x + 1} ^2} = {\left( {\sqrt {x - 2} + 1} \right)^2}\\
\Leftrightarrow x + 1 = \left( {x - 2} \right) + 2.\sqrt {x - 2} .1 + {1^2}\\
\Leftrightarrow x + 1 = x - 1 + 2\sqrt {x - 2} \\
\Leftrightarrow 2 = 2\sqrt {x - 2} \\
\Leftrightarrow \sqrt {x - 2} = 1\\
\Leftrightarrow x - 2 = 1\\
\Leftrightarrow x = 3\\
3c,\\
xP \le 10\sqrt x - 29 - \sqrt {x - 25} \,\,\,\,\,\,\,\,\,\,\left( {x \ge 25} \right)\\
\Leftrightarrow x.\dfrac{{x - 4}}{x} \le 10\sqrt x - 29 - \sqrt {x - 25} \\
\Leftrightarrow x - 4 \le 10\sqrt x - 29 - \sqrt {x - 25} \\
\Leftrightarrow x - 4 - 10\sqrt x + 29 + \sqrt {x - 25} \le 0\\
\Leftrightarrow \left( {x - 10\sqrt x + 25} \right) + \sqrt {x - 25} \le 0\\
\Leftrightarrow {\left( {\sqrt x - 5} \right)^2} + \sqrt {x - 25} \le 0\\
{\left( {\sqrt x - 5} \right)^2} \ge 0,\,\,\,\forall x \ge 25\\
\sqrt {x - 25} \ge 0,\,\,\,\forall x \ge 25\\
\Rightarrow {\left( {\sqrt x - 5} \right)^2} + \sqrt {x - 25} \ge 0,\,\,\,\,\forall x \ge 25\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {\sqrt x - 5} \right)^2} = 0\\
\sqrt {x - 25} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt x = 5\\
x - 25 = 0
\end{array} \right. \Leftrightarrow x = 25
\end{array}\)
