A= 1/6 + 1/18 + 1/36 + 1/60 + 1/90.
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\(3A=\frac{3}{2.3}+\frac{3}{6.3}+\frac{1}{12.3}+\frac{3}{20.3}+\frac{3}{30.3}+\frac{3}{42.3}\)
\(3A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(3A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{7-6}{6.7}\)
\(3A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
\(3A=1-\frac{1}{7}=\frac{6}{7}\Rightarrow A=\frac{2}{7}\)
ko phải tui ra đề đâu đề thi của trường chuyên vĩnh yên cấp 2 do sở ra đề
\(A=\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+...+\frac{1}{168}\)
\(\frac{1}{3}A=\frac{1}{54}+\frac{1}{108}+...+\frac{1}{504}\)
\(\frac{1}{3}A=\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{21.24}\)
\(=\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{21}-\frac{1}{24}\)
\(=\frac{1}{6}-\frac{1}{24}\)
\(=\frac{4-1}{24}=\frac{3}{24}=\frac{1}{8}\)
=> \(A=\frac{1}{8}:\frac{1}{3}\)\(=\frac{3}{8}\)
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
Hình như bạn viết thiếu \(\frac{1}{36}\), nếu đúng là vậy thì mình giải như sau.
Đặt A =\(\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)
3A = \(\frac{3}{3.2}+\frac{3}{3.6}+\frac{3}{3.12}+\frac{3}{3.20}+\frac{3}{3.30}+\frac{3}{3.42}\)
3A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
3A = \(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{7-6}{6.7}\)
3A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
3A = \(1-\frac{1}{7}=\frac{6}{7}\)
A = \(\frac{6}{7}:3=\frac{2}{7}\)
\(\frac{1}{6}\)+\(\frac{1}{18}\)+\(\frac{1}{36}\)+\(\frac{1}{60}\)+\(\frac{1}{90}\)\(\frac{1}{126}\)
=1/2.3+1/3.6+1/6.6+1/6.10+1/10.9+1/9.14
=1/2-1/3+1/3-1/6+1/6-1/6+1/6-1/10+1/10-1/9+1/9-1/14
=1/2-1/14
=6/14=3/7
\(\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot6}+\frac{1}{6\cdot6}+\frac{1}{6\cdot10}+\frac{1}{10\cdot9}+\frac{1}{9\cdot14}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
a) 10; 13; 18; 26; 36; 52...
c) 0; 1; 4; 9; 16; 25...
m) 1; 4; 9; 16; 25; 36; 49; 64...
p) 1; 3; 9; 27; 81; 243...
=5/18
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