toán 6 hả

X3 lên nhé

X3 lên

nhân cả tử và mẫu cho 3 là làm đc thôi

Giải:

Ta có:

\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{\left(x-3\right)\times x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x-3}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow x=3\)

De thoi ma,minh ko ghi lai de nha

=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+...+...=\(\frac{1}{6}\)

=\(\frac{1}{2}\)-\(\frac{1}{x}\)=\(\frac{1}{6}\)

Con lai bn tu lam nha . chuc bn hok tot !!!

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+....+\frac{3}{\left(x-3\right).x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x}=\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{2}-\frac{1}{6}\)

\(\Leftrightarrow\frac{1}{x}=\frac{1}{3}\)

\(\Leftrightarrow x=1\div\frac{1}{3}=3\)

Ta có : \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{\left(x-3\right)x}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{x-3}-\frac{1}{x}\)

\(=\frac{1}{2}-\frac{1}{x}\)

\(=\frac{x}{2x}-\frac{2}{2x}=\frac{x-2}{2x}\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)

\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

<=>\(\frac{1}{x+3}=\frac{1}{308}\)

=>x+3=308

<=>x=305 (nhận)

Vậy x=305

Mk sửa lại đề nha:

                \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x\left(x-2\right)}=\frac{101}{1540}\)

\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x-2}-\frac{1}{x}\right)=\frac{101}{1540}\)

\(\Leftrightarrow\)\(\frac{1}{3}-\frac{1}{x}=\frac{101}{1540}:\frac{1}{2}=\frac{101}{770}\)

\(\Leftrightarrow\)\(\frac{1}{x}=\frac{1}{3}-\frac{101}{770}=\frac{467}{2310}\)

\(\Leftrightarrow\)\(x=\frac{2310}{467}\)

P/S:  Tham khảo nhé!!!

\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left[\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right]=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right]=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\Rightarrow x+3=308\Rightarrow x=305\)

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)

=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

=> \(x+3=308\)

=> x = 305

Vậy x = 305

a)Ta có   \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=)   \(x+3=305\)=) \(x=302\)