\(\frac{1}{1\cdot3\cdot7}\)+\(\frac{1}{3\cdot7\cdot9}\)+\(\frac{1}{7\cdot9\cdot13}\)+\(\frac{1}{9\cdot13\cdot15}\)+\(\frac{1}{13\cdot15\cdot19}\)
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\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)
\(\frac{1}{1.3.7}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{3.7}\right)\)
\(\frac{1}{3.7.9}=\frac{1}{6}\left(\frac{1}{3.7}-\frac{1}{7.9}\right)\)
....
\(\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{13.15}-\frac{1}{15.19}\right)\)
Cộng các vế với nhau ta được
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+...+\frac{1}{13.15.19}=\frac{1}{6}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{37}{3.15.19}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)
\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)
\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)
\(=\frac{99.5}{486}\)
\(=\frac{495}{486}\)
Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)
\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)
\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)
\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)
\(A=66.\frac{13}{27}\)
\(A=\frac{286}{9}\)
sai hay đúng cx ko biết nha
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\)
b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\)
c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)
dễ
ta tách ra xog dùng phương pháp loại trừ đó