* Tính $\mathop {A = \lim }\limits_{x \to 0} \frac{{\sqrt {1 + 5x} - \sqrt[3]{{1 + 4x}}}}{x}$\[A = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt {1 + 5x} - 1}}{x} + \frac{{1 - \sqrt[3]{{1 + 4x}}}}{x}} \right)\]\[ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{5}{{\sqrt {1 + 5x} + 1}} - \frac{4}{{1 + \sqrt[3]{{1 + 4x}} + \sqrt[3]{{{{(1 + 4x)}^2}}}}}} \right]\]\[ = \frac{5}{2} - \frac{4}{3} = \frac{7}{6}\]
* Tính $B = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt[4]{{1 + 3x}} - \sqrt[5]{{1 + 2x}}}}{x}} \right)$
\[B = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt[4]{{1 + 3x}} - 1}}{x} + \frac{{1 - \sqrt[5]{{1 + 2x}}}}{x}} \right)=\]\[ \mathop {\lim }\limits_{x \to 0} \left[ {\frac{3}{{\sqrt[4]{{{{(1 + 3x)}^3}}} + \sqrt[4]{{{{(1 + 3x)}^2}}} + \sqrt[4]{{1 + 3x}} + 1}} + \frac{{ - 2}}{{1 + \sqrt[5]{{1 + 2x}} + \sqrt[5]{{{{(1 + 2x)}^2}}} + \sqrt[5]{{{{(1 + 2x)}^3}}} + \sqrt[5]{{{{(1 + 2x)}^4}}}}}} \right]\]\[ = \frac{3}{4} - \frac{2}{5} = \frac{7}{{20}}\]
$*\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt {1 + 5x} - \sqrt[3]{{1 + 4x}}}}{{\sqrt[4]{{1 + 3x}} - \sqrt[5]{{1 + 2x}}}}} \right) = \frac{A}{B} = \frac{{10}}{3}$