Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
\(3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{\left(x+3\right)}\right)=3\cdot\frac{49}{148}\)
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\left(x+3\right)}=\frac{147}{148}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{147}{148}\)
\(1-\frac{1}{x-1}=\frac{147}{148}\)
\(\frac{1}{x-1}=1-\frac{147}{148}\)
\(\frac{1}{x-1}=\frac{1}{148}\)
\(\Rightarrow x-1=148\)
\(\Leftrightarrow x=148+1\)
\(\Leftrightarrow x=149\)
Vậy x=149
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{49}{148}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{49}{148}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{49}{148}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{49}{148}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{49}{148}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{147}{148}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{147}{148}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{148}\)
\(\Rightarrow x+3=148\)
\(\Rightarrow x=148-3\)
\(\Rightarrow x=145\)
Vậy x = 145
_Chúc bạn học tốt_
Bạn Shinichi Kudo hình như làm sai rồi. Tại sao từ 3/x+(x+3) lại suy ra được 1/x-1/x+1.
