Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 3\sqrt 3 + 4\sqrt {12} - 5\sqrt {27} \\
= 3\sqrt 3 + 4.\sqrt {{2^2}.3} - 5.\sqrt {{3^2}.3} \\
= 3\sqrt 3 + 4.2.\sqrt 3 - 5.3.\sqrt 3 \\
= 3\sqrt 3 + 8\sqrt 3 - 15\sqrt 3 \\
= - 4\sqrt 3 \\
b,\\
B = \sqrt {32} - \sqrt {50} + \sqrt {18} \\
= \sqrt {{4^2}.2} - \sqrt {{5^2}.2} + \sqrt {{3^2}.2} \\
= 4\sqrt 2 - 5\sqrt 2 + 3\sqrt 2 \\
= 2\sqrt 2 \\
c,\\
C = \sqrt {72} + \sqrt {4\frac{1}{2}} - \sqrt {32} - \sqrt {162} \\
= \sqrt {{6^2}.2} + \sqrt {\frac{9}{2}} - \sqrt {{4^2}.2} - \sqrt {{9^2}.2} \\
= 6\sqrt 2 + \sqrt {{3^2}.\frac{1}{2}} - 4\sqrt 2 - 9\sqrt 2 \\
= - 7\sqrt 2 + 3\sqrt {\frac{1}{2}} \\
g,\\
G = \left( {\sqrt 5 + \sqrt 3 } \right)\left( {5 - \sqrt {15} } \right)\\
= \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt 5 \left( {\sqrt 5 - \sqrt 3 } \right)\\
= \sqrt 5 .\left[ {\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)} \right]\\
= \sqrt 5 .\left( {5 - 3} \right)\\
= 2\sqrt 5 \\
h,\\
H = \left( {\sqrt {32} - \sqrt {50} + \sqrt {27} } \right).\left( {\sqrt {27} + \sqrt {50} - \sqrt {32} } \right)\\
= \left( {\sqrt {{4^2}.2} - \sqrt {{5^2}.2} + \sqrt {{3^2}.3} } \right).\left( {\sqrt {{3^2}.3} + \sqrt {{5^2}.2} - \sqrt {{4^2}.2} } \right)\\
= \left( {4\sqrt 2 - 5\sqrt 2 + 3\sqrt 3 } \right)\left( {3\sqrt 3 + 5\sqrt 2 - 4\sqrt 2 } \right)\\
= \left( {3\sqrt 3 - \sqrt 2 } \right)\left( {3\sqrt 3 + \sqrt 2 } \right)\\
= {\left( {3\sqrt 3 } \right)^2} - {\sqrt 2 ^2}\\
= 27 - 2 = 25
\end{array}\)
