7x9:55
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Ta có: A = \(\frac{6}{5\times7}+\frac{6}{7\times9}+\frac{6}{9\times11}+...+\frac{6}{95\times97}+\frac{6}{97\times99}\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{95\times97}+\frac{1}{97\times99}\right)\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{6}\left(\frac{1}{5}-\frac{1}{99}\right)\)
=> A = ...
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)

Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!

917749738461936926399639748776398646491639394748947630373937366

7 - 63 = -56
7 - 63 = -56
7 - 63 = -56
nhae nhae nhae nhae
pạn



\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

\(\dfrac{2}{7}\) \(\times\) \(\dfrac{9}{7}\) - \(\dfrac{18}{49}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{18}{49}\) - \(\dfrac{18}{49}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{18}{49}\) \(\times\) 1 - \(\dfrac{18}{49}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{18}{49}\) \(\times\) ( 1 - \(\dfrac{1}{2}\) )
= \(\dfrac{18}{49}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{18}{98}\)
= \(\dfrac{9}{49}\)

\(\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{59.61}\)
\(=3\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=3\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3.56}{305}\\ =\dfrac{168}{305}\)
7×9÷55=\(\frac{63}{55}\)
7x9:55=1.145454545454545
Đây là một số thập phân được học ở lớp 5. Nhưng lớp 3 chưa thể học đến bài này vì thế có thể nó đã bị sai đề khi chỉ nằm ở lớp 3