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a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
Đặt dãy trên là A
Khi đó \(A=\frac{2}{6\times10}+\frac{2}{7\times9}+\frac{1}{64}\)
\(A=\frac{1}{30}+\frac{2}{63}+\frac{1}{64}\)
\(A=\frac{672}{20160}+\frac{640}{20160}+\frac{315}{20160}=\frac{1627}{20160}\)
\(\frac{6^2}{5\cdot7}\cdot\frac{7^2}{6\cdot8}\cdot\frac{8^2}{7\cdot9}\cdot\frac{9^2}{8\cdot10}\)
=\(\frac{6\cdot6}{5\cdot7}\cdot\frac{7\cdot7}{6\cdot8}\cdot\frac{8\cdot8}{7\cdot9}\cdot\frac{9\cdot9}{8\cdot10}\)
=\(\frac{6\cdot7\cdot8\cdot9}{5\cdot6\cdot7\cdot8}\cdot\frac{6\cdot7\cdot8\cdot9}{7\cdot8\cdot9\cdot10}\)
=\(\frac{9}{5}\cdot\frac{3}{5}\)=\(\frac{27}{25}\)
**** MIK
\(=\frac{6.6}{5.7}.\frac{7.7}{6.8}.\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{6.6.7.7.8.8.9.9}{5.6.7.8.9.10.8.7}\)
\(=\frac{27}{25}\)
Tách phần lử trên ra sao cho có thể rút gọn với phần ơn dưới
H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
Giải:
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{10}\)
\(=\dfrac{7}{30}\)
Vậy ...
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
=\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
=\(\dfrac{1}{3}-\dfrac{1}{10}\)
=\(\dfrac{7}{30}\)