\(2.A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{26.28}=\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{28-26}{26.28}\)

\(2.A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{28}=\frac{1}{2}+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{26}+\frac{1}{26}\right)-\frac{1}{28}\)

\(2.A=\frac{1}{2}-\frac{1}{28}=\frac{26}{56}=\frac{13}{28}\)=> A = \(\frac{13}{56}\)

\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)

\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)

\(A=\frac{4.1.4.1}{3.2.15.6}\)

\(A=\frac{4}{135}\)

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)

\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)

\(=\frac{5}{1}.\frac{2}{6}\)

\(=\frac{5}{1}.\frac{1}{3}\)

\(=\frac{5}{3}\)

Đọc kỹ lại đề đi Trần Cao Vỹ Lượng

= \(\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+\frac{5.5}{4.6}+\frac{6.6}{5.7}\)

= \(\frac{2.3.4.5.6}{1.2.3.4.5}+\frac{2.3.4.5.6}{3.4.5.6.7}\)

= \(\frac{2}{1}+\frac{6}{7}\)

= 2\(\frac{6}{7}\)

Mình nghĩ zậy !!!!!!!!!!!!!!!!!!

bài đó cũng có trong đề cương thi của mih

C=2.2.3.3.4.4.5.5/1.3.2.4.3.5.4.6

C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)

C=2.5/6

C=5/3

C=2^2/1.3.3^2/2.4.5^2/3.5.4^2/4.6

C= 2.2.3.3.5.5.4.4/1.3.2.4.3.5.4.6

C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)

C=5.2/6

C=5/3

\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{5^2}{3.5}.\frac{4^2}{4.6}\)

\(=\frac{2^2.3^2.5^2.4^2}{1.3.2.4.3.5.4.6}\)

\(=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}\)

\(=\frac{2.5}{6}\)

\(=\frac{10}{6}=\frac{5}{2}\)

\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+............+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(B=\frac{4^2-2^2}{\left(2.4\right)^2}+\frac{6^2-4^2}{\left(4.6\right)^2}+..........+\frac{98^2-96^2}{\left(96.98\right)^2}+\frac{100^2-98^2}{\left(98.100\right)^2}\)

\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-...............-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{100^2}\)

\(B=\frac{1}{4}-\frac{1}{10000}\)

\(B=\frac{2500}{10000}-\frac{1}{10000}\)

\(B=\frac{2499}{10000}\)

Vậy B = \(\frac{2499}{10000}\)

=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}\)

=\(\frac{1}{2}-\frac{1}{8}\)

=\(\frac{3}{8}\)

xin lỗi mình mới học lớp 5 thôi

Có cần giải tóm tắt không

chám là cái gì