Giải thích các bước giải:
Bài 4:
$\begin{array}{l}
a)\left( {2\sqrt 8 - 3\sqrt 3 + 1} \right):\sqrt 6 \\
= \left( {2.2\sqrt 2 - 3\sqrt 3 + 1} \right):\sqrt 6 \\
= \dfrac{{4\sqrt 2 - 3\sqrt 3 + 1}}{{\sqrt 6 }}\\
= \dfrac{{\sqrt 6 \left( {4\sqrt 2 - 3\sqrt 3 + 1} \right)}}{{\sqrt 6 }}\\
= \dfrac{{8\sqrt 3 - 9\sqrt 2 + \sqrt 6 }}{6}\\
b)\left( {\dfrac{1}{2}\sqrt {\dfrac{1}{2}} - \dfrac{3}{2}\sqrt {4,5} + \dfrac{2}{3}\sqrt {50} } \right):\dfrac{4}{{15}}\\
= \left( {\dfrac{1}{{2\sqrt 2 }} - \dfrac{3}{2}.\dfrac{3}{{\sqrt 2 }} + \dfrac{2}{3}.5\sqrt 2 } \right).\dfrac{{15}}{4}\\
= \left( {\dfrac{{ - 4}}{{\sqrt 2 }} + \dfrac{{10\sqrt 2 }}{3}} \right).\dfrac{{15}}{4}\\
= \dfrac{{ - 15}}{{\sqrt 2 }} + \dfrac{{25\sqrt 2 }}{2}\\
= \dfrac{{10\sqrt 2 }}{2}\\
= 5\sqrt 2
\end{array}$
Bài 5:
b) ĐKXĐ: $a,b\ge 0$
Ta có:
Áp dụng BĐT Bunhiacopski ta có:
$\begin{array}{l}
{\left( {\sqrt a + \sqrt b } \right)^2} = {\left( {1.\sqrt a + 1.\sqrt b } \right)^2} \le \left( {{1^2} + {1^2}} \right)\left( {{{\left( {\sqrt a } \right)}^2} + {{\left( {\sqrt b } \right)}^2}} \right) = 2\left( {a + b} \right)\\
\Leftrightarrow \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{4} \le \dfrac{{a + b}}{2}\\
\Leftrightarrow \dfrac{{\sqrt a + \sqrt b }}{2} \le \sqrt {\dfrac{{a + b}}{2}}
\end{array}$
Ta có đpcm.
Dấu bằng xảy ra:
$\begin{array}{l}
\Leftrightarrow \dfrac{{\sqrt a }}{1} = \dfrac{{\sqrt b }}{1}\\
\Leftrightarrow \sqrt a = \sqrt b \\
\Leftrightarrow a = b
\end{array}$
Vậy BĐT xảy ra dấu bằng khi $a=b$
