Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
b,\\
\left( {\dfrac{{\sqrt x }}{{x - 4}} + \dfrac{1}{{\sqrt x - 2}}} \right).\dfrac{{\sqrt x - 2}}{2}\\
= \left( {\dfrac{{\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \dfrac{1}{{\sqrt x - 2}}} \right).\dfrac{{\sqrt x - 2}}{2}\\
= \left( {\dfrac{{\sqrt x + \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\dfrac{{\sqrt x - 2}}{2}\\
= \dfrac{{2\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{2}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
c,\\
\left( {\dfrac{{a - 2}}{{a + 2\sqrt a }} + \dfrac{1}{{\sqrt a + 2}}} \right).\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \left( {\dfrac{{a - 2}}{{\sqrt a \left( {\sqrt a + 2} \right)}} + \dfrac{1}{{\sqrt a + 2}}} \right).\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \dfrac{{\left( {a - 2} \right) + \sqrt a }}{{\sqrt a \left( {\sqrt a + 2} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \dfrac{{a + \sqrt a - 2}}{{\sqrt a \left( {\sqrt a + 2} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right)}}{{\sqrt a \left( {\sqrt a + 2} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
2,\\
\sqrt {a + b} \le \sqrt a + \sqrt b \\
\Leftrightarrow a + b \le {\left( {\sqrt a + \sqrt b } \right)^2}\\
\Leftrightarrow a + b \le a + 2\sqrt {ab} + b\\
\Leftrightarrow 2\sqrt {ab} \ge 0,\,\,\,\,\dforall a \ge b \ge 0\\
\Rightarrow \sqrt {a + b} \le \sqrt a + \sqrt b \\
B = \sqrt {x - 5} + \sqrt {7 - x} \,\,\,\,\,\,\left( {5 \le x \le 7} \right)\\
\ge \sqrt {\left( {x - 5} \right) + \left( {7 - x} \right)} = \sqrt 2 \\
\Rightarrow {B_{\min }} = \sqrt 2 \Leftrightarrow 2\sqrt {\left( {x - 5} \right)\left( {7 - x} \right)} = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 7
\end{array} \right.
\end{array}\)
Em xem lại đề câu 1a nha!
