Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{50}{101}\)

suy ra: \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{50}{101}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{50}{101}\)

\(\frac{1}{1}-\frac{1}{x+2}=\frac{50}{101}:\frac{1}{2}=\frac{100}{101}\)

\(\frac{1}{x+2}=1-\frac{100}{101}=\frac{1}{101}\)

suy ra: \(x+2=101\)

suy ra: \(101-2=99\)

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

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

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

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

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

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

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

\(x=1\)

Mình biết làm này nhưng sợ bạn ko hiểu

2              

nhớ tk cho mk bài này như sau:

  1/2.[2/3.5+2/5.7+2/7.9+...+2/(2x+1).(2x+3)]=15/93

  1/2.[1.3-1/5+1/5-1/7+1/7-1/9+...+1/(2x+1)-1/(2x+3)]=5/31(rút gọn)

  1/3-1/(2x+3)=5/31:1/2

  1/3-1/(2x+3)=10/31

  1/2x+3=1/3-10/31

  1/2x+3=1/93

=>2x+3=93

    2x=93-3

    2x=90

      x=90:2

      x=40

 Vậy x=40

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

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

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

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

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

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

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

Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))

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

=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}=\frac{7}{15}\)

=> x = 15 (tm)

b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)

=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)

=> \(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 90

=> x = 45