Giải:
Vì tanx xác định nên \(\cos x \ne 0\). a) \(\frac{{3\sin x - 4\cos x}}{{5\sin x + 2\cos x}} = \frac{{\frac{{3\sin x - 4\cos x}}{{\cos x}}}}{{\frac{{5\sin x + 2\cos x}}{{\cos x}}}}\) \( = \frac{{3\tan x - 4}}{{5\tan x + 2}} = \frac{{3.2 - 4}}{{5.2 + 2}} = \frac{1}{6}\). b) \(\frac{{{{\sin }^3}x + 2{{\cos }^3}x}}{{2\sin x + 3\cos x}} = \frac{{\frac{{{{\sin }^3}x + 2{{\cos }^3}x}}{{{{\cos }^3}x}}}}{{\frac{{2\sin x + 3\cos x}}{{{{\cos }^3}x}}}}\) \( = \frac{{{{\tan }^3}x + 2}}{{\frac{1}{{{{\cos }^2}x}}\left( {2\tan x + 3} \right)}} = \frac{{{{\tan }^3}x + 2}}{{\left( {{{\tan }^2}x + 1} \right)\left( {2\tan x + 3} \right)}}\) \( = \frac{{{2^3} + 2}}{{\left( {{2^2} + 1} \right)\left( {2.2 + 3} \right)}} = \frac{{10}}{{5.7}} = \frac{2}{7}\).