cậu nhâ n cả 2 vế của C với 4 / 5 ý ra lu n ak

\(C=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+...+\frac{5}{101.105}\)

\(C=5.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{101.105}\right)\)

\(C=5.\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{101}-\frac{1}{105}\right)\)

\(C=\frac{5}{4}.\left(1-\frac{1}{105}\right)\)

\(C=\frac{5}{4}.\frac{104}{105}\)

\(C=\frac{26}{21}\)

\(C=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+...+\frac{5}{101.105}\)

\(\Rightarrow\frac{4}{5}C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)

\(\Rightarrow\frac{4}{5}C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{105}\)

\(\Rightarrow\frac{4}{5}C=1-\frac{1}{105}\)

\(\Rightarrow\frac{4}{5}C=\frac{104}{105}\)

\(\Rightarrow C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)

a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

      \(\frac{1}{2}.B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

         \(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)

                  \(B=\frac{100}{101}.2=\frac{200}{101}\)

b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)

      \(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

          \(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)

                 \(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)

\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)

\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=2\cdot\left(1-\frac{1}{101}\right)\)

\(=2\cdot\frac{100}{101}\)

\(=1\frac{99}{101}\)

a) \(B=\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\)

\(\Rightarrow\frac{2}{4}B=\frac{2}{4}\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\right)\)

\(\Leftrightarrow\frac{2}{4}B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)

\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)

\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{101}=\frac{100}{101}\)

\(\Leftrightarrow B=\frac{100}{101}:\frac{2}{4}=\frac{100\cdot4}{101\cdot2}=\frac{200}{101}\)

\(A=3.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)

\(A=3.\left(1-\frac{1}{105}\right)\)

\(A=3.\frac{104}{105}\)

\(A=\frac{104}{35}\)

Em yêu cầu bác nhìn xuống dưới và bác sẽ biết cách làm 

Bác thấy rồi mà còn đăng

Thay số mà làm nhé

:))

Để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của các mẫu lên tử 

Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được

tôi biết câu này nè

LÀM TẮT NHÉ :

\(C=\frac{3}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\right)\))

\(D=\frac{4}{5}\left(\frac{1}{2}-\frac{1}{7}+...+\frac{1}{102}-\frac{1}{107}\right)\)

tương tự với các phần còn lại

\(A=3\times\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)

\(A=3\times\left(1-\frac{1}{105}\right)\)

\(A=3\times\frac{104}{105}\)

\(A=\frac{104}{35}\)

\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)

\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\) 

\(4A=1-\frac{1}{x+4}\) 

\(4A=\frac{x+4-1}{x+4}\)   

\(A=\frac{x+3}{\text{4(x+4)}}\)

Bạn tự thay rồi tính nhé 

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)

\(4A=1-\frac{1}{x+4}\)

\(A=\left(1-\frac{1}{x+4}\right):4\)

Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)

A = \(\left(1-\frac{1}{16}:4\right)\)

A = \(\frac{15}{16}:4=\frac{15}{64}\)

Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)

A = \(\left(1-\frac{1}{6}\right):4\)

A \(=\frac{5}{6}:4=\frac{5}{24}\)

Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)

A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)

A = \(\frac{23}{29}:4=\frac{23}{116}\)

câu \(\frac{5}{6}\)làm thế nào

\(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+......+\frac{3}{21.25}\)

\(=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{21.25}\right)\)

\(=\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{21}-\frac{1}{25}\right)\)

\(=\frac{3}{4}\left(1-\frac{1}{25}\right)\)

\(=\frac{3}{4}.\frac{24}{25}\)

\(=\frac{18}{25}\)

\(4A=3-\frac{1}{5}+\frac{3}{5}-\frac{3}{9}+\frac{3}{9}-\frac{3}{13}+...+\frac{3}{21}-\frac{3}{25}\)\(\frac{3}{25}\)

\(4A=3-\frac{3}{25}\)

\(4A=\frac{72}{25}\)

\(A=\frac{18}{25}\)

k minh ha

\(giải:\)

\(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{21.25}\)

\(=3\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{3}{21.25}\right)\)

\(=\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{21.25}\right)\)

\(=\frac{3}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\right)\)

\(=\frac{3}{4}\left(1-\frac{1}{25}\right)\)

\(=\frac{3}{4}.\frac{24}{25}\)

\(=\frac{18}{25}\)

\(\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+...+\frac{2}{95.99}\)

\(=\frac{1}{2}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{95.99}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{95}-\frac{1}{99}\right)\)

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

\(=\frac{1}{2}.\frac{98}{99}\)

\(=\frac{49}{99}\)

Chúc cậu học tốt !!!