Đáp án:

\(\begin{array}{l}
B1:\\
a)KTM\\
b)1\\
c)6\sqrt 3  + 5\\
B2:\\
a)\left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{{3 + 8\sqrt 6 }}{{25}}\\
x = \dfrac{{3 - 8\sqrt 6 }}{{25}}
\end{array} \right.\\
c)x \in \emptyset 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a)Do:\dfrac{1}{{18}} + \dfrac{{\sqrt 2  - 2}}{{\sqrt 2 }}\\
 = \dfrac{{19 - 18\sqrt 2 }}{{18}} < 0\\
 \to KTM\\
b)\left| {2 - \sqrt 3 } \right| + \sqrt {3 - 2.\sqrt 3 .1 + 1} \\
 = 2 - \sqrt 3  + \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} \\
 = 2 - \sqrt 3  + \sqrt 3  - 1\\
 = 1\\
c)\dfrac{{4\left( {\sqrt 3  - 1} \right)}}{{3 - 1}} - \dfrac{{5\left( {\sqrt 3  + 2} \right)}}{{3 - 4}} + \dfrac{{6\left( {\sqrt 3  + 3} \right)}}{{3 - 9}}\\
 = 2\left( {\sqrt 3  - 1} \right) + 5\left( {\sqrt 3  + 2} \right) - \left( {\sqrt 3  + 3} \right)\\
 = 2\sqrt 3  - 2 + 5\sqrt 3  + 10 - \sqrt 3  - 3\\
 = 6\sqrt 3  + 5\\
B2:\\
a)\sqrt {{{\left( {x - 2} \right)}^2}}  = 3\\
 \to \left| {x - 2} \right| = 3\\
 \to \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 =  - 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 1
\end{array} \right.\\
b)DK:1 \ge x \ge  - 1\\
\sqrt {1 - x}  + \dfrac{1}{2}.\sqrt {16\left( {x + 1} \right)}  = 6\\
 \to \sqrt {1 - x}  + 2\sqrt {x + 1}  = 6\\
 \to 1 - x + 4\sqrt {\left( {x + 1} \right)\left( {1 - x} \right)}  + 4\left( {x + 1} \right) = 6\\
 \to 4\sqrt {1 - {x^2}}  + 3x + 5 = 6\\
 \to 4\sqrt {1 - {x^2}}  = 1 - 3x\\
 \to 16\left( {1 - {x^2}} \right) = 1 - 6x + 9{x^2}\\
 \to 25{x^2} - 6x - 15 = 0\\
 \to 25{x^2} - 2.5x.\dfrac{3}{5} + \dfrac{9}{{25}} - \dfrac{{384}}{{25}} = 0\\
 \to {\left( {5x - \dfrac{3}{5}} \right)^2} = \dfrac{{384}}{{25}}\\
 \to \left[ \begin{array}{l}
5x - \dfrac{3}{5} = \dfrac{{8\sqrt 6 }}{5}\\
5x - \dfrac{3}{5} =  - \dfrac{{8\sqrt 6 }}{5}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{3 + 8\sqrt 6 }}{{25}}\\
x = \dfrac{{3 - 8\sqrt 6 }}{{25}}
\end{array} \right.\\
c)DK:x \ge 1\\
3x + 1 = \sqrt {x - 1} \\
 \to 9{x^2} + 6x + 1 = x - 1\\
 \to 9{x^2} + 5x + 2 = 0\\
 \to 9{x^2} + 2.3x.\dfrac{5}{6} + \dfrac{{25}}{{36}} + \dfrac{{47}}{{36}} = 0\\
 \to {\left( {3x + \dfrac{5}{6}} \right)^2} + \dfrac{{47}}{{36}} = 0\left( {KTM} \right)\\
Do:{\left( {3x + \dfrac{5}{6}} \right)^2} + \dfrac{{47}}{{36}} > 0\forall x\\
KL:x \in \emptyset 
\end{array}\)