Tính nhanh
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56};\)
\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.......+\frac{3}{49.51};\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+.......+\frac{3}{49.51};\)
\(\frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+......+\frac{1}{2010.2011}\)
\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\)
\(=\frac{3}{8}\)
\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)
\(=1-0-0-0-...-0-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
\(d,\)giống câu a tự làm nha mỏi tay quá.
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)
=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)
=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)
1/2*3+1/3*4+1/4*5+...+1/7*8
1/2-1/3+1/3-1/4+1/4-1/5-...-1/8
1/2-1/8=3/8
1/4-1/7+1/7-1/10+1/10-1/13-...-1/52 49/52 bạn nhé
1/4-1/52=3/13
câu này mình gọi nó là S
ta có S:2=2/1*3+2/3*5+...+2/49*51
1/1-1/3+1/3-1/5+...+1/49-1/51
1/1-1/51=50/51
S=50/51*2=100/51
1/100-1/101+1/101-1/102+1/102-1/103+...+1/2010-1/2011
1/100-1/2011
bạn tích đi nhé mình còn phải đi học bạn k cho mình nhé
Câu 1:
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}=\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{8-7}{7.8}=\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
Câu 2; câu 4 làm tương tự câu 1
Câu 3
\(\frac{2A}{3}=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
Sau đó làm tương tự các câu trên
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
Sửa đề : \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}\)
\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)
\(=\frac{1}{4}-\frac{1}{52}=\frac{3}{13}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(=\frac{3}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\left(1-\frac{1}{51}\right)=\frac{3}{2}\cdot\frac{50}{51}=\frac{25}{17}\)
\(\frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+...+\frac{1}{2010.2011}\)
\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=\frac{1}{100}-\frac{1}{2011}=\frac{201}{20110}\)