\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(B=\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+...+\frac{2}{19×21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{19}-\frac{1}{21}\)

\(B=\frac{1}{3}-\frac{1}{21}\)

\(B=\frac{2}{7}\)

A=\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+...+\(\frac{1}{66}\) 

A=\(\frac{1}{1\cdot3}\) +\(\frac{1}{2\cdot3}\) +\(\frac{1}{2\cdot5}\)+...+\(\frac{1}{6\cdot11}\)

A=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{11}\)

A=\(\frac{1}{1}-\frac{1}{11}\)

=>A=\(\frac{10}{11}\)

B=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\) 

2B=\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

2B=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

2B=\(\frac{1}{3}-\frac{1}{21}\)

2B=\(\frac{2}{7}\)

B=\(\frac{2}{7}:2\)

=>B=\(\frac{1}{7}\)

\(A=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{83.85}\)

\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{83.85}\)

\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{83}-\frac{1}{85}\)

\(2A=\frac{1}{25}-\frac{1}{85}\)

\(2A=\frac{12}{425}\)

\(A=\frac{12}{425}:2\)

\(A=\frac{6}{425}\)

\(C=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(C=\frac{1}{3}-\frac{1}{21}\)

\(C=\frac{2}{7}\)

CÂU B LÀM TƯƠNG TỰ NHA 

HOK TOT

a.\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}=\frac{37}{111}-\frac{1}{111}=\frac{36}{111}=\frac{12}{37}\)

Vậy A=\(\frac{12}{37}\)

b.\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)

\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)

Vậy \(B=\frac{2}{7}\)

c.\(C=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

\(\Rightarrow C.\frac{1}{2}=\left(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\right).\frac{1}{2}\)

\(=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)

\(=\frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}\)

Vậy \(C=\frac{3}{16}\)

A = \(\frac{4}{3.7}+\frac{4}{7.9}+...+\frac{4}{107.111}\)

A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}\)=\(\frac{12}{37}\)

2 câu sau tương tự. Mik ngại làm lắm -_-

bấm vào chữ 0 đúng sẽ ra đáp án 

A = 98/303

Cơn Mưa Tình Yêu

A = 1/15 + 1/35 + 1/ 63 + 1/99 + ...+ 1/9999

A     = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)

Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)

Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101

Ax2 = 1/3 – 1/101 = 98/303

A    =  98/303 : 2

A    =  49/303

bai toan nay don

49/303

duyệt đi