1-1/3-1/65

\(A=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{63.65}\)

\(A=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{63-65}\right)\)

\(A=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{63}-\frac{1}{65}\right)\)

\(A=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)

\(A=1-\frac{62}{195}\)

\(A=\frac{133}{195}\)

Đề bài sai 

\(A=\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.100}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}\)

\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}\)

\(N=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

N=1/2x(1/3-1/5+1/5-1/7+....+1/99-1/101)

N=1/2x(1/3-1/101)

N=1/2x98/101

N=49/101

A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)

=3/2x4/3x...............x100/99

=2-1/99

=197/99

A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)

A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)

A=\(\frac{100}{2}=50\)

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%

Giải :

\(N=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

=> \(N=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

=> \(N=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\right)\right)\)

=> \(N=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\right)\)

=> \(N=\frac{98}{303}\)

N=1/2x(1/3-1/5+1/5-1/7+.....+1/99-1/101)

N=1/2x(1/3-1/101)

N=1/2x98/303

N=49/303

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{43.45}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{43.45}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\)

\(=\frac{1}{3}-\frac{1}{45}=\frac{15}{45}-\frac{1}{45}=\frac{14}{45}\)

\(\Rightarrow A=\frac{14}{45}:2=\frac{14}{90}=\frac{7}{45}\)

Vậy \(A=\frac{7}{45}\).

Áp dụng công thức : \(\frac{1}{a}-\frac{1}{a+n}=\frac{n}{a\left(a+n\right)}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{43\cdot45}\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{43}-\frac{1}{45}\right)\)

\(A=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{45}\right)\)

\(A=\frac{1}{2}\cdot\frac{14}{45}=\frac{7}{45}\)

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2.\left(1-\frac{1}{99}\right)\)

\(=2.\frac{98}{99}\)

\(=\frac{196}{99}=1\frac{97}{99}\)

Câu b sai rồi

A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\) 

A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )

A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )

A= 5. (\(1-\frac{1}{100}\))

A= 5.\(\frac{99}{100}\)

A= \(\frac{99}{20}\)

B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)

    = \(\frac{1}{2}\)-  \(\frac{1}{3}\)+\(\frac{1}{3}\)-   \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)-     \(\frac{1}{10}\)

    =  \(\frac{1}{2}\) -     \(\frac{1}{10}\)

     =       \(\frac{2}{5}\)