\(E=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}\)

Ta có \(\frac{1}{2.2}>\frac{1}{2.3}\)

\(\frac{1}{3.3}>\frac{1}{3.4}\)

...

\(\frac{1}{49.49}>\frac{1}{49.50}\)

=> \(E=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{49.49}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}=F\)

=> E > F

Xyz olm ơi . là j vậy

Số số hạng của phép tính là \(\dfrac{\left(49-1\right)}{2}+1=25\) số hạng

\(K=1x\left(100-1\right)+3x\left(100-3\right)+5x\left(100-5\right)+...+49x\left(100-49\right)=\)

\(=100x\left(1+3+5+...+49\right)-\left(1^2+3^2+5^2+...+49^2\right)=\)

Đặt

\(A=1+3+5+...+49\)

\(B=1^2+3^2+5^2+...+49^2\)

\(B=1x\left(3-2\right)+3x\left(5-2\right)+5\left(7-2\right)+...+49x\left(51-2\right)=\)

\(1x3+3x5+5x7+...+49x51-2\left(1+3+5+...+49\right)=\)

\(K=100xA-B=102xA-\left(1x3+3x5+5x7+...+49x51\right)=\)

A là cấp số cộng có 25 số hạng; d=2

Đặt

 \(C=1x3+3x5+5x7+...+49x51\)

\(6xC=1x3x\left(5+1\right)+3x5x\left(7-1\right)+5x7x\left(9-3\right)+...+49x51x\left(53-47\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9+...-47.49.51+49.51.53=\)

\(=1.3+49x51x53\Rightarrow C=\dfrac{1.3+49.51.53}{6}\)

Bạn tự tính toán nốt nhé

ai tra loi nhanh nhat se duoc k

co ca cach giai nha cac ban

\(\frac{3}{1x3}+\frac{3}{3x5}+...+\frac{3}{49x51}=\frac{3}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{49x51}\right)=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

\(S=\frac{4}{1\times3}+\frac{16}{3\times5}+\frac{36}{5\times7}+...+\frac{2500}{49\times51}\)

\(=\frac{1\times3+1}{1\times3}+\frac{3\times5+1}{3\times5}+\frac{5\times7+1}{5\times7}+...+\frac{49\times51+1}{49\times51}\)

\(=\frac{1\times3}{1\times3}+\frac{1}{1\times3}+\frac{3\times5}{3\times5}+\frac{1}{3\times5}+\frac{5\times7}{5\times7}+\frac{1}{5\times7}+...+\frac{49\times51}{49\times51}+\frac{1}{49\times51}\)

\(=1+\frac{1}{1\times3}+1+\frac{1}{3\times5}+1+\frac{1}{5\times7}+...+\frac{1}{49\times51}\) (  Có : \(\left(51-3\right)\div2+1=25\)chữ số 1 )

\(=25+\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{49\times51}\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\times\left(\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\frac{50}{51}\)

\(=25+\frac{25}{51}\)

\(=\frac{1300}{51}\)

\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)

\(=\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+...+\frac{2500}{2499}\)

\(=1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+...+1+\frac{1}{2499}\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2500}\right)\)

\(=25+\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=1-\frac{1}{51}=\frac{50}{51}\)

\(\Rightarrow S=25+\frac{50}{51}=\frac{1325}{51}\)

Vậy S=\(\frac{1325}{51}\)

Đặt \(S=\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{49\cdot51}\)

\(S=\frac{3}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

\(S=\frac{3}{2}\cdot\left(1-\frac{1}{51}\right)\)

\(\Rightarrow S=\frac{3}{2}\cdot\frac{50}{51}=\frac{3\cdot50}{2\cdot51}=\frac{150}{102}=\frac{25}{17}\)