Đáp án:
\(\begin{array}{l}
1,\\
x > \dfrac{1}{8}\\
2,\\
a,\\
Q = \dfrac{{x + 25}}{{\sqrt x }}\\
b,\\
x = 25
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
8x - 1 \ge 0\\
\sqrt {8x - 1} \ne 0
\end{array} \right. \Leftrightarrow 8x - 1 > 0 \Leftrightarrow 8x > 1 \Leftrightarrow x > \dfrac{1}{8}\\
2,\\
a,\\
Q = \left( {\dfrac{x}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }} - \dfrac{{\sqrt x + 1}}{{x - 1}}} \right).\left( {\dfrac{{x + 25}}{{x + \sqrt x + 1}}} \right)\\
= \left( {\dfrac{x}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }} - \dfrac{{\sqrt x + 1}}{{\left( {\sqrt x - 1} \right).\left( {\sqrt x + 1} \right)}}} \right).\left( {\dfrac{{x + 25}}{{x + \sqrt x + 1}}} \right)\\
= \left( {\dfrac{x}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }} - \dfrac{1}{{\sqrt x - 1}}} \right).\left( {\dfrac{{x + 25}}{{x + \sqrt x + 1}}} \right)\\
= \dfrac{{x + \left( {x + 1} \right)\left( {\sqrt x - 1} \right) - 1.\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\left( {\dfrac{{x + 25}}{{x + \sqrt x + 1}}} \right)\\
= \dfrac{{x + x\sqrt x - x + \sqrt x - 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{x\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{{{\sqrt x }^3} - {1^3}}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{x + 25}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + 25}}{{\sqrt x }}\\
b,\\
Q = 10 \Leftrightarrow \dfrac{{x + 25}}{{\sqrt x }} = 10\\
\Leftrightarrow x + 25 = 10\sqrt x \\
\Leftrightarrow x - 10\sqrt x + 25 = 0\\
\Leftrightarrow {\sqrt x ^2} - 2.\sqrt x .5 + {5^2} = 0\\
\Leftrightarrow {\left( {\sqrt x - 5} \right)^2} = 0\\
\Leftrightarrow \sqrt x - 5 = 0\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = 25
\end{array}\)
