Mik giải phía dưới rồi đó. Câu lúc nãy bạn đăng ý

\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)

\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)

\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)

\(x=\frac{4}{5}+\frac{1}{5}\)

\(x=1\)

\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)

\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)

\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)

\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)

=> x + 2 = 41 

=> x = 39

Có:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(1-0-0-0...-0-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{5}{11}:\dfrac{1}{2}=\dfrac{10}{11}\)

\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{10}{11}\)

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

\(\Rightarrow x+2=11\)

\(\Rightarrow x=11-2=9\)

Vậy x = 9.

Chúc bạn học tốt!

1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2)

= 1/2.(1/1 - 1/3) + 1/2.(1/3 - 1/5) + 1/2.(1/5 - 1/7) + ... + 1/2.(1/x -1/x+2)

= 1/2.(1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 )

= 1/2.(1/1 - 0 - 1/x+2 )

= 1/2 . ( 1/1 - 1/x+2 )

= 1/2 . ( x+2/x+2 - 1/x+2 )

= 1/2 . x+1/x+2

Mà 1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2) = 5/11

=> 1/2 . x+1/x+2 = 5/11

=> x+1/x+2 = 5/11 : 1/2

=> x+1/x+2 = 10/11

=> x+1/x+2-1 = 10/11-1

=> x+1/x+1 = 10/10

=> x + 1 = 10

=> x = 10 - 1

=> x = 9

Vậy x = 9

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)

\(\dfrac{1.2}{2.1.3}+\dfrac{1.2}{2.3.5}+\dfrac{1.2}{2.5.7}+...+\dfrac{1.2}{2x\left(x+2\right)}\dfrac{5}{11}\)\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{5}{11}\)\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)\(\dfrac{1}{2}\left(1+0+0+...+0+\dfrac{1}{x-2}\right)=\dfrac{5}{11}\)

\(\dfrac{1}{2}\left(1-\dfrac{1}{x-2}\right)=\dfrac{5}{11}\)

\(1-\dfrac{1}{x-2}=\dfrac{5}{11}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x-2}=\dfrac{10}{11}\)

\(\dfrac{1}{x-2}=1-\dfrac{10}{11}\)

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

\(\Rightarrow x-2=11\)

\(x=11+2\)

\(x=13\)

Vậy x=13

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)

\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)

\(\frac{x}{14}=\frac{16}{21}\)

\(\Rightarrow x\cdot=21=14\cdot16\)

\(\Rightarrow x\cdot21=224\)

\(\Rightarrow x=\frac{224}{21}\)

TA CÓ THỂ THẤY, VẾ TRÁI CÓ: 12 CẶP

=>   \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

<=>  \(x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)       (****)

Ta xét:    \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

=>   \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\)

=>   \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\)

=>   \(2A=1-\frac{1}{25}=\frac{24}{25}\)

=>   \(A=\frac{12}{25}\)

Ta tiếp tục xét:      \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

=>   \(3B=1+\frac{1}{3}+...+\frac{1}{3^4}\)

=>   \(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)

=>   \(2B=1-\frac{1}{3^5}=\frac{242}{243}\)

=>   \(B=\frac{121}{243}\)

THAY CÁC GIÁ TRỊ A; B VÀO PT (****) TA ĐƯỢC: 

=>   \(x+\frac{12}{25}=\frac{121}{243}\)

<=>   \(x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)

1/3.5+1/5.7+1/7.9+...+1/(2x+1)(2x+3)=5/31

1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=5/31

1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3=5/31:1/2

1/3-1/2x+3=10/31

        1/2x+3=1/3-10/31

        1/2x+3=1/63

suy ra : 2x+3=63 

              2x=63-3 

              2x=60

             x=60:2

             x=30

vay x=30

nhớ **** cho mình nha

ai tk cho mk tròn 30 với

ai tk cho mk 45 chứ ko fải 30 mk nhầm