Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
G = \sqrt {12 + 8\sqrt 2 } + \sqrt {6 - 4\sqrt 2 } \\
= \sqrt {8 + 2.2\sqrt 2 .2 + 4} + \sqrt {4 - 2.2.\sqrt 2 + 2} \\
= \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 2.2\sqrt 2 .2 + {2^2}} + \sqrt {{2^2} - 2.2.\sqrt 2 + {{\sqrt 2 }^2}} \\
= \sqrt {{{\left( {2\sqrt 2 + 2} \right)}^2}} + \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
= 2\sqrt 2 + 2 + 2 - \sqrt 2 \\
= 4 + \sqrt 2 \\
b,\\
A = \dfrac{1}{{7 + 4\sqrt 3 }} + \dfrac{1}{{7 - 4\sqrt 3 }}\\
= \dfrac{{\left( {7 - 4\sqrt 3 } \right) + \left( {7 + 4\sqrt 3 } \right)}}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}}\\
= \dfrac{{14}}{{{7^2} - {{\left( {4\sqrt 3 } \right)}^2}}} = \dfrac{{14}}{{49 - 48}} = \dfrac{{14}}{1} = 14\\
2,\\
a,\\
\sqrt {4x - 16} + \sqrt {x - 4} - \dfrac{1}{3}\sqrt {9x - 36} = 4\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 4} \right)\\
\Leftrightarrow \sqrt {4.\left( {x - 4} \right)} + \sqrt {x - 4} - \dfrac{1}{3}\sqrt {9\left( {x - 4} \right)} = 4\\
\Leftrightarrow 2\sqrt {x - 4} + \sqrt {x - 4} - \dfrac{1}{3}.3.\sqrt {x - 4} = 4\\
\Leftrightarrow 2\sqrt {x - 4} = 4\\
\Leftrightarrow \sqrt {x - 4} = 2\\
\Leftrightarrow x = 8\\
b,\\
\sqrt {{x^2} - 2x + 4} = 2x - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - 2 \ge 0\\
{x^2} - 2x + 4 = {\left( {2x - 2} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
{x^2} - 2x + 4 = 4{x^2} - 8x + 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
3{x^2} - 6x = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 2\\
c,\\
\sqrt {16{x^2} - 8x + 1} = 3\\
\Leftrightarrow \sqrt {{{\left( {4x} \right)}^2} - 2.4x.1 + 1} = 3\\
\Leftrightarrow \sqrt {{{\left( {4x - 1} \right)}^2}} = 3\\
\Leftrightarrow \left| {4x - 1} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
4x - 1 = 3\\
4x - 1 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{2}
\end{array} \right.
\end{array}\)
