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

1/3.3.[1/5.8+1/8,11+1/11.14+......+1/x.(x+3)=101/1540

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

1/3.[1/5-1/8+1/8-1/11+1/11-1/14+....+1/x-1/x+3=101/1540

1/3.[1/5-1/x+3]=101/1540

1/5-1/x+3=101/1540.3

1/5-1/x+3=303/1540

1/x+3=1/3-303/1540=1/308

=>x+3=308    =>x=305

              Vậy x=305

1/3.3(1/5.8+1/8.11+1/11.14+.....1/x(x+1)_101/1540

1/3.(1/5-1/8+1/8-1/11+1/11-1/14+....1/x+1/x+3)=101/1540

1/3.(1/5-1/x+3)=101/1540

1/5-1/x+3=101/1540/1/3=303/1540

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

x+3+308

x=305

sao lại 1/x+3=1/3-303/1540=1/308 phải là 1/x+3=1/5-303/1540=1/108

x+3=108 

x=105

giúp mik đi ạ

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

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

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)

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

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

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

\(x+3=308\)

\(\Rightarrow x=305\)

305 bạn nhé

phần a là 05 nha phần b mk ko bít

Ai làm đc giải cụ thể ra nha!thanks

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

\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

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

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

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

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

=> \(\frac{1}{x+3}=\frac{1}{308}\)

=> x + 3 = 308

=> x = 308 - 3

=> x = 305

x = 305