\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)

\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)

\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)

Xét vế phải :

\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)

\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)\(<1\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)

Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}<1\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}<1\)

=> \(A=\frac{\left(\frac{49}{1}+\frac{48}{2}+...+\frac{1}{49}\right)}{50}=\frac{49}{50.1}+\frac{48}{50.2}+...+\frac{1}{50.49}\)

=> \(A=\frac{50-1}{50.1}+\frac{50-2}{50.2}+...+\frac{50-49}{50.49}\)

=> \(A=\left(\frac{50}{50.1}+\frac{50}{50.2}+...+\frac{50}{50.49}\right)-\left(\frac{1}{50.1}+\frac{2}{50.2}+...+\frac{49}{50.49}\right)\)

=> \(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\) ( có 49 số 1/50 )

=> \(A=1+\frac{1}{2}+...+\frac{1}{49}-\frac{49}{50}=\left(1-\frac{49}{50}\right)+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\)

=> \(A=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)

Vậy A không phải là số tự nhiên 

Nếu a chia hết thì cách giải là a chia hết 1.2.....50 suy ra a chia hết cho 2,cho 3,.....,cho 50

suy ra a+2 là hợp số a chia hết 2,2chia hết cho 2

           a+3 là hợp số a chia hết cho 3, 3 chia hết cho 3

            .....................................................................

            a+ 50 là hợp số a chia hết cho 50 , 50 chia hết cho 50

Giải giúp mình nhé !!!!!!!!!!!!!

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{50\times51}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\\ A< 1-\frac{1}{51}=\frac{49}{51}\\ \Rightarrow A< 2\)