\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)

\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)

x=0.78

\(\dfrac{3^2}{8.11}+\dfrac{3^2}{11.14}+...+\dfrac{3^2}{197.200}\)

=\(3.\left(\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{197.200}\right)\)

=\(3.\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

=\(3.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

=\(3.\dfrac{3}{25}=\dfrac{9}{25}\)

Ta có: \(\dfrac{3^2}{8\cdot11}+\dfrac{3^2}{11\cdot14}+\dfrac{3^2}{14\cdot17}+...+\dfrac{3^2}{197\cdot200}\)

\(=3\left(\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+...+\dfrac{3}{197\cdot200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)

\(=3\cdot\dfrac{24}{200}=\dfrac{72}{200}=\dfrac{9}{25}\)

\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)

\(=3\cdot\frac{23}{200}\)

đúng

Đặt 3 ra ngoài

\(\Rightarrow B=3\left(\frac{3}{8.11}\right)+3\left(\frac{3}{11.14}\right)+..+3\left(\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow B=3\left(\frac{1}{8}-\frac{1}{200}\right)=3.\frac{3}{25}=\frac{9}{25}\)

Vậy \(B=\frac{9}{25}\)

Chúc bn học tốt..!

=1,798

Ta có : 

\(C=\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\)

\(C=3\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\right)\)

\(C=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(C=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(C=3.\frac{3}{25}\)

\(C=\frac{9}{25}\)

Chúc bạn học tốt ~ 

kết quả bài nãy bằng kkkkk

N=3(3/8.11 +3 /11.14 + 3/14.17 +...+3/197.200)

N=3( 1/8-1/11+1/11-1/14+1/14-1/17+...+1/197-1/200)

N=3(1/8-1/200)

N=3. 3/25=9/25

Ủng hộ mk nha

\(\Rightarrow N=\frac{9}{8.11}+\frac{9}{11.14}+\frac{9}{14.17}+....+\frac{9}{197.200}\)

\(\Rightarrow N=\frac{9}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(\Rightarrow N=3\left(\frac{1}{8}-\frac{1}{200}\right)=\frac{3}{8}-\frac{3}{200}=\frac{75}{200}-\frac{3}{200}=\frac{72}{200}=\frac{9}{25}\)

N = 3²/8.11  +  3²/11.14  +  3²/14.17  + ...+3²/197.200.

A=\(\frac{3.3}{8.11}\)+\(\frac{3.3}{11.14}\)+\(\frac{3.3}{14.17}\)+........+\(\frac{3.3}{197.200}\)

A=3\(\frac{3}{8.11}\)+3\(\frac{3}{11.14}\)+3\(\frac{3}{14.17}\)+............+3\(\frac{3}{197.200}\)

A=3.(\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+..............+\(\frac{3}{197.200}\))

A=3.(\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+.........+\(\frac{1}{197}\)-\(\frac{1}{200}\))

A=3.(\(\frac{1}{8}\)-\(\frac{1}{200}\))

A=3.(\(\frac{50}{400}\)-\(\frac{2}{200}\))

A=3.\(\frac{48}{400}\)

A=3.\(\frac{3}{25}\)

A=\(\frac{9}{25}\)