\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{97\cdot99}=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]\)

\(=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{5}{2}\left[1-\frac{1}{99}\right]\)

\(=\frac{5}{2}\cdot\frac{98}{99}=\frac{245}{99}\)

\(=\frac{5}{2}\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{5}{2}\times\frac{98}{99}\)

\(=\frac{245}{99}\)

=\(\frac{249}{99}\)

bn nhé

\(\frac{5}{11x12}+\frac{5}{12x13}+...+\frac{5}{98x99}\)

=\(\frac{5}{11}-\frac{5}{12}+\frac{5}{12}-\frac{5}{13}+...+\frac{5}{98}-\frac{5}{99}\)

=\(\frac{5}{11}-\frac{5}{99}\)

=\(\frac{40}{99}\)

Cái cuối bỏ 1 số 0 thì đúng hơn nha bạn

\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+...+\frac{5}{98.99}\)

\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{98.99}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{99}\right)\)

\(=5.\frac{8}{99}\)

\(=\frac{40}{99}\)

\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{98.99}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{99}\right)=5.\frac{8}{99}=\frac{40}{99}\)

\(E=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(\Rightarrow E=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\) (đặt  2  làm nhân tử chung để ta có các số hạng trong ngoặc có hiệu 2 số ở mẫu = tử)

\(\Rightarrow E=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow E=2.\left(1-\frac{1}{99}\right)\)

\(\Rightarrow E=2.\frac{98}{99}\)

\(\Rightarrow E=\frac{196}{99}\)

*Không biết có đúng ko :)

k roy nha

thank bạn!

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)

3747744

bằng 470 

ko có đâu mình ko rễ bị lừa :))

trả lời:

=470

hok tốt nhé

tk mk nhé

\(M=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

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

\(M=2\left(1-\frac{1}{100}\right)\)

\(M=2.\frac{99}{100}\)

\(M=\frac{99}{50}\)

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{97.99}\)

\(N=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}\left(1-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\frac{98}{99}\)

\(N=\frac{49}{33}\)

= \(\frac{1}{2}\)- \(\frac{2}{3}\)+ (\(\frac{3}{4}\)- \(\frac{3}{4}\)) + ( -\(\frac{4}{5}\)+ \(\frac{4}{5}\)) + ( \(\frac{5}{6}-\frac{5}{6}\)) - \(\frac{6}{7}\)

= \(\frac{1}{2}-\frac{2}{3}-0-0-0-\frac{6}{7}\)

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

=\(\frac{21}{42}-\frac{28}{42}-\frac{36}{42}\)

= \(\frac{-43}{42}\)

hjjjjhhbnjbbhhbbb

công chúa hoa anh đào nhắn linh tinh mình sẽ báo cáo