Giải thích các bước giải:
\(\begin{array}{l}
17,\\
a,\\
\sqrt {3 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } - \sqrt 2 \\
= \dfrac{1}{{\sqrt 2 }}.\left[ {\sqrt 2 .\left( {\sqrt {3 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } - \sqrt 2 } \right)} \right]\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 + 2\sqrt 5 } - \sqrt {6 - 2\sqrt 5 } - 2} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {5 + 2.\sqrt 5 + 1} - \sqrt {5 - 2\sqrt 5 + 1} - 2} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} - 2} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 + 1 - \sqrt 5 + 1 - 2} \right)\\
= 0\\
b,\\
\sqrt {4 - \sqrt 7 } - \sqrt {4 + \sqrt 7 } + \sqrt 7 \\
= \dfrac{1}{{\sqrt 2 }}.\left[ {\sqrt 2 .\left( {\sqrt {4 - \sqrt 7 } - \sqrt {4 + \sqrt 7 } + \sqrt 7 } \right)} \right]\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {8 - 2\sqrt 7 } - \sqrt {8 + 2\sqrt 7 } + \sqrt {14} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} + \sqrt {14} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 7 - 1 - \sqrt 7 - 1 + \sqrt {14} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {14} - 2} \right)\\
= \sqrt 7 - \sqrt 2 \\
c,\\
\sqrt {6,5 + \sqrt {12} } + \sqrt {6,5 - \sqrt {12} } + 2\sqrt 6 \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {13 + 2\sqrt {12} } + \sqrt {13 - 2\sqrt {12} } } \right) + 2\sqrt 6 \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt {12} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {12} - 1} \right)}^2}} } \right) + 2\sqrt 6 \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {12} + 1 + \sqrt {12} - 1} \right) + 2\sqrt 6 \\
= \dfrac{1}{{\sqrt 2 }}.2\sqrt {12} + 2\sqrt 6 \\
= 2\sqrt 6 + 2\sqrt 6 \\
= 4\sqrt 6 \\
18,\\
a,\\
\sqrt {5{x^2}} = 2x + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
2x + 1 \ge 0\\
5{x^2} = {\left( {2x + 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - \dfrac{1}{2}\\
{x^2} - 4x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - \dfrac{1}{2}\\
\left[ \begin{array}{l}
x = 2 + \sqrt 5 \\
x = 2 - \sqrt 5
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 2 \pm \sqrt 5 \\
b,\\
\dfrac{{\sqrt {2x - 3} }}{{\sqrt {x - 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge \dfrac{3}{2}} \right)\\
\Leftrightarrow \dfrac{{2x - 3}}{{x - 1}} = 4\\
\Leftrightarrow 2x - 3 = 4x - 4\\
\Leftrightarrow 2x = 1\\
\Leftrightarrow x = \dfrac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,\,x \ge \dfrac{3}{2}} \right)\\
c,\\
1 + \sqrt {3x + 1} = 3x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - \dfrac{1}{3}} \right)\\
\Leftrightarrow \sqrt {3x + 1} = 3x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
3x - 1 \ge 0\\
3x + 1 = {\left( {3x - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
9{x^2} - 9x = 0
\end{array} \right. \Leftrightarrow x = 1
\end{array}\)
