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

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

\(4A=1-\frac{1}{x+4}\) 

\(4A=\frac{x+4-1}{x+4}\)   

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

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

\(giải:\)

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

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

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

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

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

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

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

\(x+\frac{2}{5.9}+\frac{2}{9.13}+\frac{2}{13.17}+...+\frac{2}{41.45}=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{2}{4}\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\frac{8}{45}\right]=\frac{-37}{45}\)

\(\Leftrightarrow x+\frac{4}{45}=\frac{-37}{45}\)

\(\Leftrightarrow x=-\frac{41}{45}\)

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

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

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

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

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 

TẬP HỢP RA HAI NHÓM .MỘT NHÓM SỐ ÂM.CÒN NHÓM KIA LÀ SỐ DƯƠNG MÀ TÍNH

                               STUDY   WELL

      K NHA

MK XIN CẢM ƠN CÁC BẠN NHÌU

C = 24.7 −35.9 +27.10 −39.13 +...+2301.304 −3401.405 

\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)

 \(C=\frac{25}{152}-\frac{4}{27}\)

\(C=\frac{67}{4104}\)

Study well 

\(C=\frac{2}{4\cdot7}-\frac{3}{5\cdot9}+\frac{2}{7\cdot10}-\frac{3}{9\cdot13}+...+\frac{2}{301\cdot304}-\frac{3}{401\cdot405}\)

\(C=\left(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{301\cdot304}\right)-\left(\frac{3}{5\cdot9}+\frac{3}{9\cdot13}+...+\frac{3}{401\cdot405}\right)\)

\(C=\frac{2}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{301\cdot304}\right)-\frac{3}{4}\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{401\cdot405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{2}{3}\cdot\frac{77}{304}-\frac{3}{4}\cdot\frac{82}{405}\)

\(C=\frac{77}{456}-\frac{41}{270}\)

\(C=\frac{349}{20520}\)

Không chắc =))

\(M=\frac{1}{3.4}+\frac{7}{3.4}+\frac{2}{3.5}+\frac{14}{5.9}-\frac{4}{9.13}=\frac{8}{3.4}+\frac{2}{3.5}+\frac{2}{9}\left(\frac{7}{5}-\frac{2}{13}\right)\)

=> \(M=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{9}.\frac{81}{5.13}=\frac{2}{3}\left(1+\frac{1}{5}\right)+\frac{18}{5.13}\)

=> \(M=\frac{2}{3}.\frac{6}{5}+\frac{18}{5.13}=\frac{4}{5}+\frac{18}{5.13}=\frac{2}{5}\left(2+\frac{9}{13}\right)=\frac{2}{5}.\frac{35}{13}\)

=> \(M=\frac{14}{13}\)

Ta có : \(x+\frac{1}{1.5}+x+\frac{1}{5.9}+x+\frac{1}{9.13}+......+x+\frac{1}{397.401}=101x\)

\(\Leftrightarrow\left(x+x+x+......+x\right)+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)

\(\Leftrightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)

\(\Rightarrow x=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\)

\(\Rightarrow4x=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{397.401}\)

\(\Rightarrow4x=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{397}-\frac{1}{401}\)

\(\Rightarrow4x=1-\frac{1}{401}\)

\(\Rightarrow4x=\frac{400}{401}\)

\(\Rightarrow x=\frac{400}{401}.\frac{1}{4}=\frac{100}{401}\)

tui biết giải, mà k biết có bao nhiêu x, bạn tính sao ra 100x vậy bạn?

Đơn giải thôi bạn chỉ cần lấy công thức tính số số hạng là ra thôi

(397 - 1) : 4 + 1 = 100 (số)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(P=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{197.201}\)
\(P=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{197.201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}+\frac{1}{13}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{201}{201}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\frac{200}{201}\)
\(P=\frac{50}{67}\)
 Vậy \(P=\frac{50}{67}\)

\(P=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{197\cdot201}\)

\(=3\cdot\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{197\cdot201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{197\cdot201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{201-1}{201}\right)\)

\(=\frac{3}{4}\cdot\frac{200}{201}\)

\(\Rightarrow B=\frac{50}{67}\)

\(\frac{4}{3}\) x P = \(\frac{4}{1x5}\)+ \(\frac{4}{5x9}\)+ ...... + \(\frac{4}{197x201}\)

\(\frac{4}{3}\)x P = \(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{197}-\frac{1}{201}\)

\(\frac{4}{3}\)x P = 1 - \(\frac{1}{201}\)

\(\frac{4}{3}\)x P = \(\frac{200}{201}\)

P = \(\frac{200}{201}\) : \(\frac{4}{3}\)= \(\frac{50}{67}\)