\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{17\cdot21}< 1\)

\(A=\dfrac{4}{4}\cdot\left(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+...+\dfrac{1}{17\cdot21}\right)< 1\)

\(A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{17}-\dfrac{1}{21}< 1\)

\(A=1-\dfrac{1}{21}< 1\) (đúng) (đpcm).

Đề sai

\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)

`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`

`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`

`= 1/5 - 1/25`

`= 4/25`

\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{44}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\dfrac{8}{45}=-\dfrac{37}{45}\\ x=-\dfrac{37}{45}-\dfrac{8}{45}\\ x=-1\)

bạn học nhà cô Phương à

a) \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-...-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-\left(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\) \(\Rightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-2.\dfrac{3}{16}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}-\dfrac{3}{8}=\dfrac{5}{8}\\ \Rightarrow\dfrac{x}{2008}=\dfrac{5}{8}+\dfrac{3}{8}\\ \Rightarrow\dfrac{x}{2008}=1\\ \Rightarrow x=2008\)

b) \(\dfrac{7}{x}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{7}{15}\\ \Rightarrow x=15\)

c) \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{15}{93}\)

\(\Rightarrow2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}\right)=\dfrac{15}{93}.2\)

\(\Rightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\\ \Rightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\\ \Rightarrow\dfrac{2x}{3\left(2x+3\right)}=\dfrac{10}{31}\\ \Rightarrow\dfrac{10.3\left(2x+3\right)}{31}=2x\\ \Rightarrow\dfrac{30\left(2x+3\right)}{31}=2x\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{31}:2\\ \Rightarrow x=\dfrac{30\left(2x+3\right)}{62}\\ \Rightarrow x=\dfrac{15\left(2x+3\right)}{31}\\\Rightarrow\dfrac{15\left(2x+3\right)}{x}=31\\ \Rightarrow\dfrac{30x+45}{x}=31\\ \Rightarrow30+\dfrac{45}{x}=31\\ \Rightarrow \dfrac{45}{x}=1\\ \Rightarrow x=45\)

a/ \(\dfrac{x}{2008}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-............-\dfrac{1}{120}=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{1}{10}+\dfrac{1}{15}+.......+\dfrac{1}{120}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\left(\dfrac{2}{20}+\dfrac{2}{30}+.......+\dfrac{2}{240}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{15.16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{15}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}-\dfrac{3}{16}=\dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{x}{2008}=\dfrac{13}{16}\)

\(\Leftrightarrow x=1631,5\)

Vậy ..................

Tính chất của phân số bạn cần biết như sau:

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Gọi biểu thức trên là A ,ta có:

\(A=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+\dfrac{1}{17\cdot21}+\dfrac{1}{21\cdot25}\)

\(4A=\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+\dfrac{4}{17\cdot21}+\dfrac{4}{21\cdot25}\)

\(4A=\dfrac{9-5}{5\cdot9}+\dfrac{13-9}{9-13}+\dfrac{17-13}{13\cdot17}+\dfrac{21-17}{17\cdot21}+\dfrac{25-21}{21\cdot25}\)

Áp dụng tính chất phân số đã nêu ở trên, ta được:

\(4A=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{25}\)

\(4A=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{5}{25}-\dfrac{1}{25}=\dfrac{4}{25}\)

\(A=4A:4=\dfrac{4}{25}:4=\dfrac{16}{25}\)

Vậy \(A=\dfrac{16}{25}\)

\(-\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}...+\dfrac{4}{n\left(n+4\right)}\right)\) \(=-\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{n}+\dfrac{1}{n+4}\right)=-\left(1-\dfrac{1}{n+4}\right)=-1+\dfrac{1}{n+4}\)

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

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

\(\Rightarrow x=\dfrac{100}{401}\)

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)

`#3107`

a)

\(\dfrac{11}{12}-\left(\dfrac{2}{5}+\dfrac{3}{4}x\right)=\dfrac{2}{3}?\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{11}{12}-\dfrac{2}{3}\\ \Rightarrow\dfrac{2}{5}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{2}{5}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{3}{20}\\ \Rightarrow x=-\dfrac{3}{20}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{5}\)

Vậy, \(x=-\dfrac{1}{5}\)

b)

\(\dfrac{-2}{5}+\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}-\dfrac{-2}{5}\\ \Rightarrow\dfrac{5}{3}\cdot\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{30}\div\dfrac{5}{3}\\ \Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\\ \Rightarrow\dfrac{4}{15}x=\dfrac{3}{2}-\left(-\dfrac{23}{50}\right)\\ \Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\\ \Rightarrow x=\dfrac{147}{20}\)

Vậy, \(x=\dfrac{147}{20}\)

c)

\(\dfrac{1}{2}+\dfrac{3}{4}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{1}{4}\div\dfrac{3}{4}\\ \Rightarrow x=-\dfrac{1}{3}\)

Vậy, \(x=-\dfrac{1}{3}.\)

\(#Emyeu1aithatroi...\)

(2/5 + 3/4 . x)= 11/12 -2/3

(2/5 +3/4 . x)= 1/4

3/4 . x          = 1/4 - 2/5

3/4 . x          = -3/20

x                  = -3/20 : 3/4

x                  = -1/5

Vậy .....