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\(\left[9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
có 9 số 1 có 9 số hạng
\(=\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=1\)
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}=1+\frac{24.19}{2}=229\)
Đặt
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{10^2}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}........\frac{-99}{100}\)
Ta có: \(-A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{99}{100}=\frac{3.8.15....99}{4.9.16......100}\)
Đễ dễ rút go5nta viết tử và mẫu của -A dưới dạng tích các số tự nhiên liên tiếp:
\(-A=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)......\left(9.11\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)........\left(10.10\right)}=\frac{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}{\left(2.3.4......10\right).\left(2.3.4.....10\right)}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}\)
\(=>A=-\frac{11}{20}\)
Vậy..........
\(=\left(1-\frac{1}{10}\right).\left(1-\frac{2}{10}\right)...\left(1-\frac{10}{10}\right)...\left(1-\frac{15}{10}\right)=\left(1-\frac{1}{10}\right).\left(1-\frac{2}{10}\right)...0...\left(1-\frac{15}{10}\right)=0\)