\(A=0,2113727891\)

\(\frac{1}{6}=0,166666667\)

\(\frac{1}{4}=0,25\)

\(\Leftrightarrow\frac{1}{6}< A< \frac{1}{4}\)

Đặt A = 1 - 1/2² - 1/3² - 1/4² - ....... - 1/10²

=> A>1-1/2.3 - 1/3.4 - 1/4.5 - ........ - 1/10.11

=> A> 1 - (1/2.3 + 1/3.4 + 1/4.5 + ..... + 1/10.11)

=> A> 1 - (1/2 -1/3 +1/3 - 1/4 + 1/4 -1/5+...+1/10-1/11)

=> A> 1 - (1/2 - 1/11)

=> A> 1 - 9/22

mà 9/22  < 1  nên (1 - 9/22) : dương

=> (1/9/22) > 0

=> A>0 (điều phải chứng minh)

\(\frac{1}{2^2}>\frac{1}{1.2};\frac{1}{3^2}>\frac{1}{2.3};.....;\frac{1}{10^2}>\frac{1}{9.10}\)

\(\Rightarrow1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{10^2}>1-\frac{1}{1.2}-\frac{1}{2.3}-....-\frac{1}{9.10}\)

\(\Rightarrow1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{10^2}>1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-...-\left(\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{10^2}>1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-....-\frac{1}{9}+\frac{1}{10}=\frac{1}{10}>0\)

=>ĐPCM

A= 1 - ( \(\frac{1}{2^2}\)+  \(\frac{1}{3^2}\)+  \(\frac{1}{4^2}\)+...+  \(\frac{1}{10^2}\))
          ________________________________
                                     B
Vì \(\frac{1}{2^2}\)<  \(\frac{1}{1.2}\);....; \(\frac{1}{10^2}\)<  \(\frac{1}{9.10}\)
=> B < \(\frac{1}{1.2}\)+  \(\frac{1}{2.3}\)+...+  \(\frac{1}{9}\)-  \(\frac{1}{10}\)
=> B < \(\frac{1}{1}\)-  \(\frac{1}{2}\)+  \(\frac{1}{2}\)-  \(\frac{1}{3}\)+...+  \(\frac{1}{9}\)-  \(\frac{1}{10}\)= \(\frac{1}{1}\)-  \(\frac{1}{10}\)= \(\frac{9}{10}\)
=> A = 1-B > 1 - \(\frac{9}{10}\)> 0
=> A > 0 (  \(Đpcm\))
 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}< 1\)

\(\Rightarrowđpcm\)

Ta co: \(\frac{1}{2^2}< \frac{1}{1.2}\)

           \(\frac{1}{3^2}< \frac{1}{2.3}\)

             ..................

         \(\frac{1}{10^2}< \frac{1}{9.10}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow A< 1-\frac{1}{10}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

Ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

          \(\frac{1}{3^3}=\frac{1}{3.3}< \frac{1}{2.3}\)

                ........

      \(\frac{1}{10^2}=\frac{1}{10.10}< \frac{1}{9.10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< 1-\frac{1}{10}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< 1\)

Vậy...

\(\frac{x}{7}=\frac{x+1}{14}\Leftrightarrow14x=7x+7\Leftrightarrow7x=7\Leftrightarrow x=1\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)

\(\Leftrightarrow1\le x\le6\Leftrightarrow x=1;2;3;4;5;6\)

\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)

\(\Leftrightarrow0\le x\le5\Leftrightarrow x=0;1;2;3;4;5\)

\(\frac{x}{7}=\frac{x+1}{14}\)

=> \(\frac{x\cdot2}{7\cdot2}=\frac{x+1}{14}\)

=> \(2x=x+1\)

=> \(2x-x-1=0\)

=> \(1x-1=0\)

=> \(x=1\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)

=> \(1\le x\le6\)

=> \(x=\left\{1;2;3;4;5;6\right\}\)

\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)

=> \(0\le x\le5\)

=> \(x=\left\{0;1;2;3;4;5\right\}\)

Cảm ơn các bạn nhé !!!