a, Ta có : \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\)

\(\Rightarrow\frac{13}{38}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)

b, Ta có: \(3^{301}>3^{300}=\left(3^3\right)^{100}=27^{100}\left(1\right)\)

               \(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\left(2\right)\)

 Do \(25^{100}< 27^{100}\Rightarrow5^{200}< 3^{300}\)\(\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow5^{199}< 5^{200}< 3^{300}< 3^{301}\Rightarrow5^{199}< 3^{301}\)

c, Ta có: \(\frac{10^{2018}+5}{10^{2018}-8}=\frac{10^{2018}-8+13}{10^{2018}-8}=1+\frac{13}{10^{2018}-8}\)

               \(\frac{10^{2019}+5}{10^{2019}-8}=\frac{10^{2019}-8+13}{10^{2019}-8}=1+\frac{13}{10^{2019}-8}\)

Do \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\Rightarrow\frac{10^{2018}+5}{10^{2018}-8}>\frac{10^{2019}+5}{10^{2019}-8}\)

a

nAK.DNX. 0pwi9dOjkciopjopoijasd

có mà nguyễn thị thối

A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

A < 1 - \(\frac{1.}{100}\)

A < \(\frac{99}{100}< \frac{199}{100}\)

=> A < \(\frac{199}{100}\)

b,

S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)

S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)

S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)

S = \(\frac{1.11}{2.10}\)

S = \(\frac{11}{20}\)

Ta có: 

\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+\frac{11^3}{30}-\frac{13^3}{42}+\frac{15^3}{56}-\frac{17^3}{72}+...+\frac{199^3}{9900}\)

\(=3^2.\left(1+\frac{1}{2}\right)-5^2.\left(\frac{1}{2}+\frac{1}{3}\right)+7^2.\left(\frac{1}{3}+\frac{1}{4}\right)-9^2.\left(\frac{1}{4}+\frac{1}{5}\right)+...+199^2.\left(\frac{1}{99}+\frac{1}{100}\right)\)

\(=3^2+\left(\frac{3^2}{2}-\frac{5^2}{2}\right)-\left(\frac{5^2}{3}-\frac{7^2}{3}\right)+\left(\frac{7^2}{4}-\frac{9^2}{4}\right)-\left(\frac{9^2}{5}-\frac{11^2}{5}\right)+...+\left(\frac{197^2}{99}-\frac{199^2}{99}\right)+\frac{199^2}{100}\)

\(=3^2-8+8-8+...+8+\frac{199^2}{100}=3^2+\frac{199^2}{100}< 3^2+\frac{199.200}{100}=9+398=407\)

\(\Rightarrow A< 407.2=814\)