a) Biến đổi vế trái ta có:
\( VT = \dfrac{3}{2}\sqrt 6+ 2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\)
\(=3\dfrac{\sqrt 6}{2}+2\dfrac{\sqrt{2}}{\sqrt 3}-4\dfrac{\sqrt 3}{\sqrt 2}\)
\(=3\dfrac{\sqrt 6}{2}+2\dfrac{\sqrt 2\sqrt 3}{\sqrt 3 .\sqrt 3}-4.\dfrac{\sqrt 3 .\sqrt 2}{\sqrt 2.\sqrt 2}\)
\(=3\dfrac{\sqrt 6}{2}+2\dfrac{\sqrt 6}{3}-4\dfrac{\sqrt 6}{2}\)
\(=3\dfrac{\sqrt 6 .3}{2.3}+2\dfrac{\sqrt 6 .2}{3.2}-4\dfrac{\sqrt 6 .3}{2.3}\)
\(=9\dfrac{\sqrt 6}{6}+4\dfrac{\sqrt 6}{6}-12\dfrac{\sqrt 6}{6}\)
\(=(9+4-12)\dfrac{\sqrt 6}{6}=\dfrac{\sqrt 6}{6}=VP\).
b) Biến đổi vế trái ta có:
\(VT = \left( {x\sqrt {\dfrac{6}{x}} + \sqrt {\dfrac{2x}{3}} + \sqrt {6x} } \right):\sqrt {6x} \)
\(\eqalign{
& = \left( {x\sqrt {{{6x} \over {{x^2}}}} + \sqrt {{{2x.3} \over {{3^2}}}} + \sqrt {6x} } \right):\sqrt {6x} \cr
& = \left( {x{{\sqrt {6x} } \over {\sqrt {{x^2}} }} + {{\sqrt {6x} } \over {\sqrt {{3^2}} }} + \sqrt {6x} } \right):\sqrt {6x} \cr
& = \left( {x{{\sqrt {6x} } \over x} + {{\sqrt {6x} } \over 3} + \sqrt {6x} } \right):\sqrt {6x} \cr
& = \left( {1.\sqrt {6x} + {1 \over 3}\sqrt {6x} + \sqrt {6x} } \right):\sqrt {6x} \cr
& = \left( {1 + {1 \over 3} + 1} \right)\sqrt {6x} :\sqrt {6x} \cr
& = {7 \over 3}\sqrt {6x} :\sqrt {6x} \cr
& = {7 \over 3}\sqrt {6x} .{1 \over {\sqrt {6x} }} = \dfrac{7}{3} =VP.\cr} \)
