Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)

\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)

\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)

Mà \(\frac{1}{9}>\frac{1}{10}\)

\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)

Vậy : M > N

\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{x}=\dfrac{127}{256}\)

Đặt VT là A

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{x}\right)=\dfrac{127}{256}\)

\(\Leftrightarrow A=1-\dfrac{1}{x}=\dfrac{127}{256}\)

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{129}{256}\)

\(\Rightarrow x=\dfrac{256}{129}\)

\(A=\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)\left(\frac{1}{36}-1\right)...\left(\frac{1}{100}-1\right)\)

\(-A=\left(1-\frac{1}{16}\right)\left(1-\frac{1}{25}\right)\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{100}\right)\)

\(-A=\frac{15}{16}\cdot\frac{24}{25}\cdot\frac{35}{36}\cdot...\cdot\frac{99}{100}\)

\(-A=\frac{\left(3\cdot5\right)\left(4\cdot6\right)\left(5\cdot7\right)...\left(9\cdot11\right)}{\left(4\cdot4\right)\left(5\cdot5\right)\left(6\cdot6\right)...\left(10\cdot10\right)}\)

\(-A=\frac{\left(3\cdot4\cdot5\cdot...\cdot9\right)\left(5\cdot6\cdot7\cdot...\cdot11\right)}{\left(4\cdot5\cdot6\cdot...\cdot10\right)\left(4\cdot5\cdot6\cdot...\cdot10\right)}\)

\(-A=\frac{3\cdot11}{10\cdot4}=\frac{33}{40}\)

\(A=-\frac{33}{40}\)

\(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A< \frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)

\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< \frac{1}{2}-\frac{1}{100}=\frac{49}{100}=\left(\frac{7}{10}\right)^2\)

Ta có \(\frac{25}{36}=\left(\frac{5}{6}\right)^2\)

Ta thấy \(\frac{5}{6}=\frac{25}{30}>\frac{7}{10}=\frac{21}{30}\Rightarrow\left(\frac{7}{10}\right)^2< \left(\frac{5}{6}\right)^2\Rightarrow A< \left(\frac{7}{10}\right)^2< \left(\frac{5}{6}\right)^2=\frac{25}{36}\)

Câu này á ???

\(A=1+\frac{1}{2}+...+\frac{1}{16}\)

= \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{12}\right)+\left(\frac{1}{13}+...+\frac{1}{16}\right)\)

> \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)

=\(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

=\(1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)

= \(1+2\times\frac{13}{12}\)

= \(1+\frac{13}{6}\)

= \(1+2+\frac{1}{6}\)

= \(3+\frac{1}{6}\)>\(3\)

=> \(A>3+\frac{1}{6}>3\)

=> \(A>3+\frac{1}{6}>B\)

=> \(A>B\)