\(a)\)

\(( x + 1 ) + ( x + 2 ) + ..... + ( x + 100 ) = 5750\)

\(\rightarrow \underbrace{x + x + ..... + x }_{ \text{100 lần x } }\) \(+ ( 1 + 2 + ..... + 100 ) = 5750\)

\(\rightarrow 100x + \dfrac{[( 100 - 1 ) : 1 + 1 ]( 100 + 1 )}{2} = 5750\)

\(\rightarrow 100x + 5050 = 5750\)

\(\rightarrow 100x = 700\)

\(\rightarrow x = 7\)

\(\text{ Vậy }\) \(x = 7 \)

\(b)\)

\(\text{ Ta có :}\)

\(\dfrac{1}{2^2} < \dfrac{1}{1.2}\)

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

\(\dfrac{1}{4^2} < \dfrac{1}{3.4}\)

\(.......\)

\(\dfrac{1}{2021^2} < \dfrac{1}{2020.2021}\)

\(\rightarrow B = \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ......... + \dfrac{1}{2021^2} \) \(\rightarrow B < \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + ........ \dfrac{1}{2020.2021}\)

\(\rightarrow B < 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ....... + \dfrac{1}{2020} - \dfrac{1}{2021}\)

\(\rightarrow B < 1 - \dfrac{1}{2021}\)

\(\rightarrow B < 1\)

\(\text{ Công thức tổng quát :}\)

\( \dfrac{1}{2^2} + ....... + \dfrac{1}{n^2} \) \(< \) \(\dfrac{1}{1.2} + \dfrac{1}{2.3} + ..... + \dfrac{1}{(n-1)n}\)