tìm x biết ( 5/1.4 + 5/4.7 + ... + 5/97.100)
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A=5(1/1.4+1/4.7+1/7.10+...+1/97.100)
A:5.3=3/1.4+3/4.7+...+3/97.100
A.3/5=1-1/4+1/4+--1/7+...+1/94-1/97+1/97-1/100
A.3/5=1-1/100
A.3/5=99/100
A=99/100:3/5
A=33/20

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\)\(\frac{1}{132}\)= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)\(=1-\frac{1}{12}=\frac{11}{12}\)

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`
`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`
`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`
`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`
`S_1 = 5/3 (1 - 1/100)`
`S_1 = 5/3 . 99/100`
`S_1 = 33/20`

\(\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)=\frac{0,33x}{2009}\)
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)
\(\left(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
\(1-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\Rightarrow2009x99=0,33x\times100\)
198891:100:0,33=6027=x

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{97\cdot100}=\frac{0,33\cdot x}{2009}\cdot3\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,99\cdot x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,99x}{2009}\)
\(\frac{99}{100}=\frac{0,99x}{2009}\)
=>0,99x*100=2009*99
99x=2009*99
=>x=2009
Vậy x=2009
\(0,33\cdot\frac{x}{2009}\) hay \(\frac{0,33\cdot x}{2009}\)

3A/5=3/1.4+3/4.7+3/7.10+...+3/97.100
3A/5=(4-1)/1.4+(7-4)/4.7+....+(100-97)/97.100
3A/5=1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100
3A/5=1-1/100=99/100
A=99/100:3/5=33/20
Vậy A=33/20