=2/10+3/10+4/10+......+13/10

=\(\frac{2+3+4+......+13}{10}\)

=90/10=9

k cho mình nha

=9 nhe cac ban

=1/10+1/10+3/10+4/10+5/10+6/10+7/10+8/10+9/10

=1/10+45/10

=46/10=23/5

\(\begin{array}{l}\left( {\frac{{20}}{7}.\frac{{ - 4}}{{ - 5}}} \right) + \left( {\frac{{20}}{7}.\frac{3}{{ - 5}}} \right) = \frac{{20}}{7}.\left( {\frac{{ - 4}}{{ - 5}} + \frac{3}{{ - 5}}} \right)\\ = \frac{{20}}{7}.\left( {\frac{{ - 1}}{{ - 5}}} \right) = \frac{{20}}{7}.\frac{1}{5} = \frac{{20}}{{35}} = \frac{4}{7}\end{array}\)

Đặt \(x=20+\frac{800}{20+\frac{800}{20+\frac{800}{20+\frac{800}{20+.....}}}}\)

Ta thấy mẫu số của số hạng thứ hai trong biểu thức của x là 20 + ... lại cũng là x

Vậy:

\(x=20+\frac{800}{x}\)

=> \(x^2-20x-800=0\)

\(x_1=40;x_2=-20\left(loại\right)\)

=> \(x=40\)

ĐS: 40

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)

\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}-\frac{1}{100}=\frac{25}{100}-\frac{1}{100}=\frac{24}{100}=\frac{6}{25}\)

\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{9}\right)=1+\frac{1}{2}-\frac{1}{3}=1\frac{1}{6}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

         \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

           \(=1-\frac{1}{100}\)

            \(=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)

\(B=1-\frac{1}{100}=\frac{99}{100}\)

~ Hok tốt ~