Đáp án:
$\begin{array}{l}
1)a)2\sqrt {45} + \sqrt 5 - 3\sqrt {80} \\
= 2.3\sqrt 5 + \sqrt 5 - 3.4\sqrt 5 \\
= 6\sqrt 5 + \sqrt 5 - 12\sqrt 5 \\
= - 5\sqrt 5 \\
b)\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + \dfrac{2}{{\sqrt 3 + 1}} - 6\sqrt {\dfrac{{16}}{3}} \\
= 2 - \sqrt 3 + \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} - 6.\dfrac{4}{{\sqrt 3 }}\\
= 2 - \sqrt 3 + \sqrt 3 - 1 - 2\sqrt 3 .4\\
= 1 - 8\sqrt 3 \\
c){\tan ^2}{40^0}.{\sin ^2}{50^0} - 3 + \left( {1 - \sin {{40}^0}} \right).\left( {1 + \sin {{40}^0}} \right)\\
= {\cot ^2}{50^0}.{\sin ^2}{50^0} - 3 + 1 - {\sin ^2}{40^0}\\
= \dfrac{{{{\cos }^2}{{50}^0}}}{{{{\sin }^2}{{50}^0}}}.{\sin ^2}{50^0} - 3 + {\cos ^2}{40^0}\\
= {\cos ^2}{50^0} + {\cos ^2}{40^0} - 3\\
= {\sin ^2}{40^0} + {\cos ^2}{40^0} - 3\\
= 1 - 3\\
= - 2\\
2)a)Dkxd:4 - 3x \ge 0 \Leftrightarrow x \le \dfrac{4}{3}\\
\sqrt {4 - 3x} = 8\\
\Leftrightarrow 4 - 3x = 64\\
\Leftrightarrow 3x = - 60\\
\Leftrightarrow x = - 20\\
Vậy\,x = - 20\\
b)Dkxd:x \ge 2\\
\sqrt {4x - 8} - 12\sqrt {\dfrac{{x - 2}}{9}} = - 1\\
\Leftrightarrow 2\sqrt {x - 2} - 12.\dfrac{{\sqrt {x - 2} }}{3} = - 1\\
\Leftrightarrow 2\sqrt {x - 2} - 4\sqrt {x - 2} = - 1\\
\Leftrightarrow - 2\sqrt {x - 2} = - 1\\
\Leftrightarrow \sqrt {x - 2} = \dfrac{1}{2}\\
\Leftrightarrow x - 2 = \dfrac{1}{4}\\
\Leftrightarrow x = \dfrac{9}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{9}{4}\\
c)DKxd:x \ge 0\\
\left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) = 7\\
\Leftrightarrow 2x - 4\sqrt x + \sqrt x - 2 - 7 = 0\\
\Leftrightarrow 2x - 3\sqrt x - 9 = 0\\
\Leftrightarrow 2x - 6\sqrt x + 3\sqrt x - 9 = 0\\
\Leftrightarrow \left( {\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right) = 0\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow x = 9\left( {tmdk} \right)\\
Vậy\,x = 9
\end{array}$
