ta có

(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)

=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)

19/36+17/90+1/12

=(19/36+1/12)+17/90

=7/12+17/90

=105/180+34/180

=139/180

1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45

=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5

=1/1-1/3+1/3-1/2....+1/9-1/5

=1/1

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17+18 = 171 nhé

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17+18.  = 171

A= 1/2 >1/10 

1/2<1/10

có kết quả r mà

\(=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{15}=\dfrac{6}{60}+\dfrac{5}{60}+\dfrac{4}{60}=\dfrac{15}{60}=\dfrac{1}{4}\)

\(=\dfrac{1.6}{10.6}-\dfrac{\left(-1\right).5}{12.5}+\dfrac{1.4}{15.4}=\dfrac{6}{60}+\dfrac{-5}{60}+\dfrac{4}{60}=\dfrac{5}{60}=\dfrac{1}{12}\)

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2007/2009

<=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2007/2009

<=> 2[(1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ... + 1/x - 1/(x+1)] = 1 - 2/2009

<=> 2[1/2 - 1/(x+1)] = 2(1/2 - 1/2009)

<=> x+1 = 2009

<=> x = 2008

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