Đáp án:
\(\begin{array}{l}a)\,\,3\sqrt 3 \\b)\,\,4\sqrt 5 \end{array}\)
Giải thích các bước giải:
a) \(7\sqrt {12} - 5\sqrt {20} - \frac{1}{3}\sqrt {27} \)
\(\begin{array}{l} = 7\sqrt {{2^2}.3} - 5\sqrt {{2^2}.5} - \frac{1}{3}\sqrt {{3^2}.3} \\ = 7.2\sqrt 3 - 5.2\sqrt 3 - \frac{1}{3}.3\sqrt 3 \\ = 14\sqrt 3 - 10\sqrt 3 - \sqrt 3 \\ = \left( {14 - 10 - 1} \right)\sqrt 3 \\ = 3\sqrt 3 \end{array}\)
b) \(\sqrt {{{\left( {6 - 3\sqrt 5 } \right)}^2}} + \sqrt {41 + 12\sqrt 5 } \)
\(\begin{array}{l} = \left| {6 - 3\sqrt 5 } \right| + \sqrt {{6^2} + 2.6.\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} \\ = 3\sqrt 5 - 6 + \sqrt {{{\left( {6 + \sqrt 5 } \right)}^2}} \,\,\left( {Do\,\,6 - 3\sqrt 5 < 0} \right)\\ = 3\sqrt 5 - 6 + \left| {6 + \sqrt 5 } \right|\\ = 3\sqrt 5 - 6 + 6 + \sqrt 5 \,\,\left( {Do\,\,6 + \sqrt 5 > 0} \right)\\ = 4\sqrt 5 \end{array}\)
