Đáp án:
\(\begin{array}{l}
C1:\\
a) - 3\\
b)1\\
C2:\\
a)x = 17\\
b)\left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - \dfrac{2}{3}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a)\left( {3.\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{3}.3\sqrt 3 - 4\sqrt 3 + \sqrt 3 } \right).\sqrt 3 \\
= \left( {\sqrt 3 + \sqrt 3 - 4\sqrt 3 + \sqrt 3 } \right).\sqrt 3 \\
= - \sqrt 3 .\sqrt 3 = - 3\\
b)\dfrac{{12\left( {3 + \sqrt 3 } \right)}}{{9 - 3}} + \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - \dfrac{{3\left( {2 + \sqrt 3 } \right)}}{{4 - 3}}\\
= 2\left( {3 + \sqrt 3 } \right) + \sqrt 3 + 1 - 3\left( {2 + \sqrt 3 } \right)\\
= 6 + 2\sqrt 3 + \sqrt 3 + 1 - 6 - 3\sqrt 3 \\
= 1\\
C2:\\
a)DK:x \ge 1\\
\sqrt {36\left( {x - 1} \right)} - \sqrt {x - 1} = 12 + \sqrt {4\left( {x - 1} \right)} \\
\to 6\sqrt {x - 1} - \sqrt {x - 1} - 2\sqrt {x - 1} = 12\\
\to 3\sqrt {x - 1} = 12\\
\to \sqrt {x - 1} = 4\\
\to x - 1 = 16\\
\to x = 17\\
b)\sqrt {{{\left( {3x - 1} \right)}^2}} = 3\\
\to \left| {3x - 1} \right| = 3\\
\to \left[ \begin{array}{l}
3x - 1 = 3\\
3x - 1 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - \dfrac{2}{3}
\end{array} \right.
\end{array}\)
