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Ta có phần tử \(=\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\)
\(=\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+\left(\frac{19}{1}+1\right)-19\)
\(=\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{1}+\frac{20}{20}-20\)
\(=20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}\right)\left(1\right)\)
Thay (1) vào P ta được :
\(P=\frac{20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}}\)
\(=20\)
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+10}\)
\(=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\cdot\frac{9}{22}\)
\(=\frac{9}{11}\)
cộng hết tất cả 1/1+2+3+.....+10 thì ta chỉ cần cộng 1+2+3+4+5+6+7+8+9+10 là xong rồi tự tính
A = \(1+\frac{1}{\left(2\times3\right):2}+\frac{1}{\left(3\times4\right):2}+....+\frac{1}{\left(19\times20\right):2}\)
A = \(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+....+\frac{2}{19\times20}\)
A = \(2\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)=2\times\left(\frac{1}{1}-\frac{1}{20}\right)=\frac{19}{10}\)
Bài này dễ ẹc mà !