biến đổi được : \(\frac{\left(x-1\right)\left(x-1\right)-\left(x+1\right)\left(x+1\right)+4}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)

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

Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)

\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)

 \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)

\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)

\(=\frac{1.41}{39.3}\)

\(=\frac{41}{117}\)

Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}\)

\(=\frac{\left(1.2.3...38\right)\left(4.5.6...61\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)

B = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)

B = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

B = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2017}\right)\)

B = \(\frac{1}{2}\left(\frac{2017}{2017}-\frac{1}{2017}\right)\)

B = \(\frac{1}{2}.\frac{2016}{2017}\)

B = \(\frac{1008}{2017}\)

Vậy B = \(\frac{1008}{2017}\)

Chúc bạn học tốt . Có bài gì khó mik sẽ giúp bạn ( Chỉ toán 6 hoặc 7 trở xuống thui đó )

\(B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2015\cdot2017}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2015\cdot2017}\right)\)

\(B=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\left(1-\frac{1}{2017}\right)\)

\(B=\frac{1}{2}\cdot\frac{2016}{2017}\)

\(B=\frac{1008}{2017}\)

\(B=\frac{1}{1.3}+\frac{1}{3.7}+...+\frac{1}{2005.2007}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2015.2017}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\)

\(2B=1-\frac{1}{2017}\)

\(2B=\frac{2016}{2017}\)

\(B=\frac{2016}{2017}:2\)

\(B=\frac{1008}{2017}\)

Vậy  \(B=\frac{1008}{2017}\)

\(1-\frac{3}{2.10}-\frac{3}{4.15}-\frac{3}{6.20}-\frac{3}{8.25}-...-\frac{3}{198.500}\)

\(=1-\left(\frac{3}{2.10}+\frac{3}{4.15}+\frac{3}{6.20}+...+\frac{3}{198.500}\right)\)

  \(=1-\frac{3}{2.5}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=1-\frac{3}{10}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=1-\frac{3}{10}.\left(1-\frac{1}{100}\right)\)

\(=1-\frac{3}{10}.\frac{99}{100}\)

\(=1-\frac{297}{1000}\)

\(=\frac{703}{1000}\)

P/s : Không biết đúng hông nha, làm đại

\(\frac{5}{a}+\frac{3}{a+4}=\frac{5.\left(a+4\right)+3a}{a.\left(a+4\right)}=\frac{5a+20+3a}{a^2+4a}\)

                          \(=\frac{8a+20}{a^2+4a}\)

\(\frac{4}{c-5}+\frac{2}{2c+3}\) \(=\frac{4\left(2c+3\right)+2\left(c-5\right)}{\left(c-5\right)\left(2c+3\right)}\)

                                        \(=\frac{8c+12+2c-10}{2c^2+3c-10c-15}\)

                                         \(=\frac{10c-2}{2c^2-7c-15}\)

câu còn lại tương tự nha

mk phải đi học rồi

\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)

\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)

\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)

\(=\frac{5}{7}.0=0\)

c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)

\(=\frac{27}{5}.10=27.2=54\)

\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)

\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)

cho mk hỏi, bn có thể lm thêm c d dc ko