\(A=\dfrac{1}{50}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{1}{50}-\dfrac{5}{14}\)

\(=\dfrac{14-250}{700}=\dfrac{-236}{700}=-\dfrac{59}{175}\)

\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)

\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)

\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)

\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)

\(=-\dfrac{69}{45}\)

\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)

a) \(\left(\dfrac{11}{12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)

\(\Rightarrow\left(\dfrac{11}{1.12}+\dfrac{11}{12.23}+\dfrac{11}{23.34}+...+\dfrac{11}{89.100}\right)+x=\dfrac{5}{3}\)

\(\Rightarrow\left(1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}\right)+x=\dfrac{5}{3}\)

\(\Rightarrow1-\dfrac{1}{100}+x=\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{5}{3}-1+\dfrac{1}{100}\)

\(\Rightarrow x=\dfrac{500}{300}-\dfrac{300}{300}+\dfrac{3}{300}\)

\(\Rightarrow x=\dfrac{203}{300}\)

b) \(\left(\dfrac{5}{11.16}+\dfrac{5}{16.21}+...+\dfrac{5}{19.24}\right)-x+\dfrac{1}{3}=\dfrac{7}{3}\)

=>\(\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{19}-\dfrac{1}{24}\right)-x=\dfrac{7}{3}-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{1}{11}-\dfrac{1}{24}-x=2\)

\(\Rightarrow-x=2-\dfrac{1}{11}+\dfrac{1}{24}\)

\(\Rightarrow-x=\dfrac{528}{264}-\dfrac{24}{264}+\dfrac{11}{264}\)

\(\Rightarrow x=\dfrac{515}{264}\)

c) Câu hỏi của Đàm Chu Hữu An - Toán lớp 6 - Học toán với OnlineMath

\(\dfrac{9}{8}-\dfrac{1}{2}-\dfrac{1}{6}-...........-\dfrac{1}{72}\)

\(=\dfrac{9}{8}-\left(\dfrac{1}{2}+\dfrac{1}{6}+..........+\dfrac{1}{72}\right)\)

\(=\dfrac{9}{8}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{8.9}\right)\)

\(=\dfrac{9}{8}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{9}{8}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{9}{8}-\dfrac{8}{9}\)

\(=\dfrac{17}{72}\)

2=1.2

6=2.3

12=3.4

20=4.5

30=5.6

.....

tụ hỉu rồi nhỉ

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)

\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\)

\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\right)\)

\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< \dfrac{5}{6}\left(2\right)\)

\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)

\(\rightarrowđpcm\)

Chúc bạn học tốt!

\(=5\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)\)

\(=5\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)

\(=5\cdot\dfrac{5}{24}=\dfrac{25}{24}\)

`5/12 + 5/20 + 5/30 + 5/42 + 5/56`

`= 5/(3.4) + 5/(4.5) + 5/(5.6) + 5/(6.7) + 5/(7.8)`

`= 5 . (1/(3.4) + 1/(4.5) + ... + 5/(7.8))`

`= 5 . (1/3 - 1/4 + 1/4 - 1/5 + ...+ 1/7 - 1/8)`

`= 5 . (1/3 - 1/8) `

`= 5 . 5/24 = 25/24`