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\(\frac{1}{337.291}+\frac{583}{291}-\frac{2}{337.291}=\frac{583}{291}-\frac{1}{337.291}=\frac{583.337-1}{337.291}=\frac{169652}{98067}\)
\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)
\(\frac{\left(1.3.2.4.3.5......\left(n-2\right)\left(n\right)\left(n-1\right)\left(n+1\right)\right)}{2.2.3.3.4.4...n.n}=\frac{\left(n+1\right)}{2.n}\)
nhầm dấu, ngoặc đầu tiên là dấu trừ chứ không phải dấu cộng
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
Chúc bạn học tốt !!!
Trả lời:
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\frac{8}{9}\)
\(=0\)
\(\frac{84}{-83}=\frac{-84}{83}\)
\(\frac{-84}{83}+\frac{1}{83}=-1\)
\(\frac{-337}{331}+\frac{6}{331}=-1\)
\(\frac{1}{83}< \frac{6}{331}\Rightarrow\frac{84}{-83}>\frac{-337}{331}\)
Xin lỗi mink mới học lớp 5 năm nay lên lớp 6 mink ko thể giúp bn đc sorry nha!
1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)