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\(a;343-4\times\left(x+1\right)=143\)
\(4\times\left(x+1\right)=343-143\)
\(4\times\left(x+1\right)=200\)
\(x+1=200:4\)
\(x+1=50\)
\(x=50-1=49\)
Mấy câu hỏi trẻ trâu thì tự tính nhé
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy x=305
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(3.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{4620}\)
\(\frac{1}{x+3}=...\) (tự làm tiếp)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+1\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+1}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)
\(\frac{1}{x+1}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=> x + 1 = 308
=> x = 308 - 1
=> x = 307
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{5}{1540}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
Vậy x = 305
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{5}{1540}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
Vậy x = 305
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+........+\(\frac{1}{x.\left(x+3\right)}\)=\(\frac{101}{1540}\)
3(.\(\frac{1}{5.8}+\frac{1}{8.11}\)+\(\frac{1}{11.14}+.......+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>\(x+3=308\)
\(x=308-3=305\)
Vậy \(x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 5
x = 303
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(x-5\right)+3\left(2-3x\right)=5x+7\)
\(\Leftrightarrow2x-10+6-9x=5x+7\)
\(\Leftrightarrow\left(2x-9x\right)+\left(6-10\right)=5x+7\)
\(\Leftrightarrow-7x-4=5x+7\)
\(\Leftrightarrow-7x-5x=4+7\)
\(\Leftrightarrow-12x=11\)
\(\Leftrightarrow x=\frac{-11}{12}\)
\(3x-5\left(x-2\right)+7=4x-12\)
\(\Leftrightarrow3x-5x-10+7=4x-12\)
\(\Leftrightarrow\left(3x-5x\right)-\left(10-7\right)=4x-12\)
\(\Leftrightarrow-2x-3=4x-12\)
\(\Leftrightarrow-2x-4x=3-12\)
\(\Leftrightarrow-6x=-15\)
\(\Leftrightarrow x=\frac{-15}{-6}=\frac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^
duyệt đi
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{5}+\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\left(\frac{1}{5}-\frac{1}{x+3}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{x}-\frac{1}{x}\right)=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>x+3=308
x=308-3
x=305
Vậy x=305