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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 :
\(P=\frac{20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}}\)
\(=20\)
Mình ghi lời giải luôn
=\(\frac{1}{\frac{2x3}{2}}+\frac{1}{\frac{3x4}{2}}+...+\frac{1}{\frac{19x20}{2}}\)
=\(\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{19x20}\)
=\(2x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
=2x(1/2-1/20)
=2x9/20
=9/10
Thôi chết, mình quên chưa cộng với 1
Bạn cộng 1 vào nữa
=19/10
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
ở dãy 1 thì số đứng sau bằng tổng hai số đứng trước
ta có 5 số tiếp theo la 40,74, 136,...
\(a.\)\(1\frac{2}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+5\frac{3}{7}\)
\(=\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{3}\cdot\frac{3}{2}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(=\frac{4}{2}+\frac{38}{7}\)
\(=2+\frac{38}{7}\)
\(=\frac{14}{7}+\frac{38}{7}\)
\(=\frac{52}{7}\)
\(b.1\frac{1}{3}-1\frac{1}{4}:1\frac{1}{2}+2\frac{3}{4}\cdot3\frac{2}{3}\)
\(=\frac{4}{3}-\frac{5}{4}:\frac{3}{2}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{4}\cdot\frac{2}{3}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{6}+\frac{121}{12}\)
\(=\frac{16}{12}-\frac{10}{12}+\frac{121}{12}\)
\(=\frac{6}{12}+\frac{121}{12}\)
\(=\frac{127}{12}\)
\(c.7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{2}{5}\cdot\frac{2}{1}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{14}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{70}{15}-\frac{12}{15}-\frac{10}{15}\)
\(=\frac{58}{15}-\frac{10}{15}\)
\(=\frac{48}{15}=\frac{16}{5}\)
\(\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(2+\frac{38}{7}\)
\(\frac{52}{7}\)
khó quá