\(\Leftrightarrow\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{202}{1540}\)

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

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

\(\Leftrightarrow x+3=308\)

\(\Leftrightarrow x=305\)

Vậy x=305

.....

<=>\(\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)

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

<=> x+3=308

<=> x=305

\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+..+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)

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

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

Vạy x = 305

\(\dfrac{2}{40}+\dfrac{2}{88}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{202}{1540}\)

\(\Leftrightarrow2\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{202}{1540}\)

\(\Leftrightarrow\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

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

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

\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)

Ta có 2/40 + 2/88 + 2/154 + ... + 2/x( x + 3) = 202

=> 2/5 x 8 + 2/8 x 11 + ... + 2/x( x + 3 ) = 202

=> 1/5 x 8 + 1/8 x 11 + ... + 1/x( x + 3 ) = 202 : 2

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

=> 1/5 - 1/x + 3 = 101

=> 1/x + 3 = 1/5 - 101

=> 1/X + 3 = 504/5

=> 504(x + 3 ) = 5

Có ai kết bạn vs tui ko

\(\frac{2}{40}+\frac{2}{888}+\frac{2}{154}+......+\frac{2}{x.\left(x+3\right)}=202\)

\(=\frac{2}{5.8}+\frac{2}{888}+\frac{2}{11.14}+............+\frac{2}{x.\left(x+3\right)}=202\)

=>Đề sai  bạn sửa lại nha

Chúc bạn học tốt

\(\dfrac{1}{x^2+7x+10}+\dfrac{1}{x^2+13x+40}+\dfrac{1}{x^2+19x+88}+\dfrac{1}{x^2+25x+154}\)

\(=\dfrac{1}{\left(x+2\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+8\right)}+\dfrac{1}{\left(x+8\right)\left(x+11\right)}+\dfrac{1}{\left(x+11\right)\left(x+14\right)}\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+11}+\dfrac{1}{x+11}-\dfrac{1}{x+14}\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}\)