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Xét :\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+..+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Khi đó ,phương trình cần giải có dạng \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\)
Giờ thì việc tìm x quá đơn giản đúng không!
Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
Bạn tham khảo
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{101}{1504}\)
\(\Rightarrow\frac{1}{3}\cdot\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{101}{1504}\)
\(\Rightarrow\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)
\(\Rightarrow\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}\cdot\frac{3}{1}=\frac{303}{1504}\)
- Đến đây tự tính nhé :v
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (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
\(\frac{1}{5.8} + \frac{1}{8.11} + ..+ \frac{1}{x.(x+3)}=\frac{101}{1540} \)
Nhân với 3 vào cả hai vế ta có:
\(\frac{3}{5.8} + \frac{3}{8.11} + ..+ \frac{3}{x.(x+3)}=\frac{303}{1540} \)
\(\frac{8-5}{5.8} + \frac{11-8}{8.11} + ..+ \frac{(x+3)-x}{x.(x+3)}=\frac{303}{1540} \)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3\)
\(x=305\)
Em cũng làm đễn đấy mà quên nhân 3 với \(\frac{101}{1540}\) nên cô dạy đội bồi dưỡng cho sai ( mà cũng hên là vô được :)) )
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{101}{1540}.3\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
\(\frac{1}{5\times8}+\frac{1}{8\times11}+...+\frac{1}{x\times\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}\times\left(\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{x\times\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}\div\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}\times3\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3\)
x = 305
Chúc bạn học tốt ^^
\(\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\frac{1}{5}-\frac{1}{3}-\frac{1}{x+3}=\frac{101}{1540}\)
\(\frac{1}{15}-\frac{1}{x+3}=\frac{101}{1540}\)
\(\frac{1}{x+3}=\frac{1}{15}-\frac{101}{1540}\)
\(\frac{1}{x+3}=\frac{1}{924}\)
=> x = 924 -3
=> x = 921
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 \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Leftrightarrow x=308-3\)
\(\Leftrightarrow x=305\)
Vậy \(x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1504}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1504}\)
\(\frac{1}{x+3}=-\frac{11}{7502}\)
\(x+3=\left(7502.1\right):\left(-11\right)\)
\(x+3=7502:\left(-11\right)\)
\(x+3=-682\)
\(x=-682-3\)
\(x=-385\)
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