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

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x=2009

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

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)\(=1-\frac{1}{12}=\frac{11}{12}\)

giải hộ mình bài về công nhân với

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+.....+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)

= \(2.\dfrac{99}{100}\)

=\(\dfrac{99}{50}\)

câu 6 dễ mà câu 7

A=3-3/4+3/4-3/7+3/7-3/10+...+3/94-3/97+3/97-3/100

A=3-3/100

A=300/100-3/100

A=297/100

Câu 6:

\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)

=>\(\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=-3+3=0\)

=>\(\frac{x+2010}{2005}+\frac{x+2010}{2006}+\frac{x+2010}{2007}=0\)

=>x+2010=0

=>x=-2010

Câu 7:

\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\cdots+\frac{3}{97\cdot100}\)

\(=1-\frac14+\frac14-\frac17+\cdots+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

`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`

E=3434

\(\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)=\frac{0,33x}{2009}\)

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)

\(\left(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}\right)=\frac{0,33x}{2009}\)

\(1-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{2009}\Rightarrow2009x99=0,33x\times100\)

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 \(\frac{0,33\cdot x}{2009}\)