đề sai ak bn

mình sửa lại đề nhé

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

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

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

\(\Rightarrow\)  \(\frac{1}{3}\)(\(\frac{1}{5}\)- \(\frac{1}{x+3}\)) = \(\frac{101}{1540}\)

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

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

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

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

\(\Rightarrow\) x+3 = 308

\(\Rightarrow\)x        = 305

Vậy x = 305

a)1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540

<=>1/3(3/5.8+3/8.11+...+3/x(x+3)     =101/1540

<=>1/3(1/5-1/8+1/8-1/11+...+1/x-1/x+3=101/1540

<=>1/5-1/x+3=303/1540<=>1/x+3=1/308

<=>x+3=308<=>x=305

Nguồn CHTT, hihi !

Tham gia event này đi mọi người https://olm.vn/hoi-dap/detail/227766827875.html

\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right) \)
                                                               \(\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Không có gtri A=? ak bạn??

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

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

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)

=\(\frac{101}{1540}\)

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

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

\(\Rightarrow\)x+3=308

\(\Rightarrow\)x=308-3=305

b)Mk chưa nghĩ ra

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2}{9}\)

\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{x+1-6}{6\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{x-5}{6x+6}=\frac{1}{9}\)

\(\Rightarrow9x-45=6x+6\)

\(\Rightarrow3x=51\)

\(\Rightarrow x=17\)

Vậy x = 17

      \(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)

\(\Leftrightarrow x+3=103\)

\(\Leftrightarrow x\)\(=103-3\)

\(\Leftrightarrow x\)\(=100\)

Vậy x = 100

~~~~~~~Hok tốt~~~~~~~~

ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)

\(\Rightarrow x+3=103\)

\(\Rightarrow x=100\)

nhớ k nha

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14+}+...+\frac{1}{x\left(x+3+\right)}=\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{3\left(x+3\right)}=3.\frac{98}{1545}\)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)

\(\Leftrightarrow x+3=103\)

\(\Leftrightarrow x+103-3\)

\(\Leftrightarrow x=100\)

vậy x=100

~~~~~~~~~~~~~~~~~~~~hk tốt~~~~~~~~~~~~~~~~~~~~

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)

b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)

\(=\frac{5}{4}.\frac{4n}{12n+9}\)

\(=\frac{5n}{12n+9}\)

( sai đề )

a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)

                                                              \(=1-\frac{1}{32}=\frac{31}{32}\)

b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)

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

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

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

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{4620}\)

\(\frac{1}{x+3}=\frac{823}{4620}\)

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

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

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

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

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

aloalo