Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = 1 + \sqrt {9 + 6\sqrt 2 } \\
x = 1 - \sqrt {9 + 6\sqrt 2 }
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = 2\sqrt 2 - \sqrt 3 + 2\\
x = - 2\sqrt 2 + \sqrt 3 + 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt {{{\left( {x - 1} \right)}^2}} = \sqrt {9 + 6\sqrt 2 } \\
\to {x^2} - 2x + 1 = 9 + 6\sqrt 2 \\
\to {x^2} - 2x - 8 - 6\sqrt 2 = 0\\
\Delta ' = 1 + 8 + 6\sqrt 2 = 9 + 6\sqrt 2 \\
\to \left[ \begin{array}{l}
x = 1 + \sqrt {9 + 6\sqrt 2 } \\
x = 1 - \sqrt {9 + 6\sqrt 2 }
\end{array} \right.\\
b)\sqrt {{{\left( {x - 2} \right)}^2}} = \left| {\sqrt 3 - \sqrt 8 } \right|\\
\to \left| {x - 2} \right| = 2\sqrt 2 - \sqrt 3 \\
\to \left[ \begin{array}{l}
x - 2 = 2\sqrt 2 - \sqrt 3 \\
x - 2 = - 2\sqrt 2 + \sqrt 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\sqrt 2 - \sqrt 3 + 2\\
x = - 2\sqrt 2 + \sqrt 3 + 2
\end{array} \right.
\end{array}\)
