13: \(\lim_{x\to+\infty}\left(\sqrt[3]{x^3+4x^2}-x\right)\)

\(=\lim_{x\to+\infty}\frac{x^3+4x^2-x^3}{\sqrt[3]{\left(x^3+4x^2\right)^2}+x\cdot\sqrt[3]{x^3+4x^2}+x^2}\)

\(=\lim_{x\to+\infty}\frac{4x^2}{x^2\cdot\sqrt[3]{\left(1+\frac{4}{x}\right)^2}+x^2\cdot\sqrt[3]{1+\frac{4}{x}}+x^2}\)

\(=\lim_{x\to+\infty}\frac{4}{\sqrt[3]{\left(1+\frac{4}{x}\right)^2}+\sqrt[3]{1+\frac{4}{x}}+1}=\frac{4}{\sqrt[3]{\left(1+0\right)^2}+\sqrt[3]{1+0}+1}=\frac43\)

12: \(\lim_{x\to2}\frac{4-x^2}{x^3-8}\)

\(=\lim_{x\to2}\frac{\left(2-x\right)\left(2+x\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\lim_{x\to2}\frac{-\left(x+2\right)}{x^2+2x+4}=\frac{-\left(2+2\right)}{2^2+2\cdot2+4}=-\frac{4}{4+4+4}=-\frac{4}{12}=-\frac13\)

11: \(\lim_{x\to4}\frac{4x-1}{x^2-8x+16}=\lim_{x\to4}\frac{4x-1}{\left(x-4\right)^2}=+\infty\)

vì \(\lim_{x\to4}4x-1=4\cdot4-1=15>0;\left(x-4\right)^2=\left(4-4\right)^2=0\)

a)

\(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(=\dfrac{146}{13}-\left(\dfrac{18}{7}+\dfrac{68}{13}\right)\)

\(=\dfrac{146}{13}-\dfrac{18}{7}-\dfrac{68}{13}\)

\(=\left(\dfrac{146}{13}-\dfrac{68}{13}\right)-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

b)

\(\dfrac{2}{7}\times5\dfrac{1}{4}-\dfrac{2}{7}\times3\dfrac{1}{4}\)

\(=\dfrac{2}{7}\times\dfrac{21}{4}-\dfrac{2}{7}\times\dfrac{13}{4}\)

\(=\dfrac{2}{7}\times\left(\dfrac{21}{4}-\dfrac{13}{4}\right)\)

\(=\dfrac{2}{7}\times2\)

\(=\dfrac{4}{7}\)

\(a,11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(=\dfrac{146}{13}-\left(\dfrac{18}{7}+\dfrac{68}{13}\right)\)

\(=\dfrac{146}{13}-\dfrac{68}{13}-\dfrac{18}{7}\)

\(=6-\dfrac{18}{7}\)

\(=\dfrac{24}{7}\)

\(b,\dfrac{2}{7}\times5\dfrac{1}{4}-\dfrac{2}{7}\times3\dfrac{1}{4}\)

\(=\dfrac{2}{7}\times\dfrac{21}{4}-\dfrac{2}{7}\times\dfrac{13}{4}\)

\(=\dfrac{2}{7}\times\left(\dfrac{21}{4}-\dfrac{13}{4}\right)\)

\(=\dfrac{2}{7}\times2\)

\(=\dfrac{4}{7}\)

a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)

= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)

= - 1 + 1  - \(\dfrac{11}{20}\)

=   0 - \(\dfrac{11}{20}\)

= - \(\dfrac{11}{20}\)

b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)

= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)

= \(\dfrac{10}{12}\)

= \(\dfrac{5}{6}\)

c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)

= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)

= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)

= \(\dfrac{14}{3}\)

Đáp số là 1442/1287 nhé

=\(\dfrac{1442}{1287}\)

a) = 1.1204
b) = 0.5714

a) = 1.1204
b) = 0.5714

    1/9 x 4/11 + 3/4 +7/11 x 1/9

= 1/9 x (4/11 + 7/11) + 3/4

= 1/9 x 1 + 3/4

=1/9 + 3/4

= 31/36