(1-1/3).(1-1/6).(1-1/10)...(1-1/780) (ko có dấu "...") 
Giải: (1-1/3).(1-1/6).(1-1/10)...(1-1/780) 
= 2/3 . 5/6....779/780 
= 4/6 . 10/12.....1558/1560 
= 1.4 . 2.5 .... 38.41/ 2.3 . 3. 4. .....39.40 
= ( 1.2.3....38).(4.5....41)/(2.3.4....39)(3... 
triệt tiêu xong còn 41/39.3= 41/117 
ĐS = 41/117

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

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

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

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

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

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

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

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

Cảm ơn mn ạ.

1.

7/15:(1/2-9/10xX)-1/6=13/6

=>7/15:(1/2-9/10xX)=13/6+1/6

=>7/16:(1/2-9/10xX)=7/3

=>1/2-9/10xX=7/16:7/3

=>1/2-9/10xX=3/16

=>9/10xX=1/2-3/16

=>9/10xX=5/16

=>X=5/16:9/10

=>X=25/72

Bài 1 

a; \(\dfrac{7}{19}\) x \(\dfrac{1}{3}\) + \(\dfrac{7}{19}\) x \(\dfrac{2}{3}\)

= \(\dfrac{7}{19}\) x (\(\dfrac{1}{3}+\dfrac{2}{3}\))

= \(\dfrac{7}{19}\) x 1

= \(\dfrac{7}{19}\)

b; 15 x \(\dfrac{2121}{4343}\) + 15 x \(\dfrac{212121}{434343}\)

= 15 x \(\dfrac{21}{43}\) + 15 x \(\dfrac{21}{43}\)

= 15 x \(\dfrac{21}{43}\) x (1 + 1)

= 15 x \(\dfrac{21}{43}\) x 2

= (15 x 2) x \(\dfrac{21}{43}\)

= 30 x \(\dfrac{21}{43}\)

= \(\dfrac{630}{43}\)

C; \(\dfrac{1991}{1990}\) x \(\dfrac{1992}{1991}\) x \(\dfrac{1993}{1992}\) x \(\dfrac{1994}{1993}\) x \(\dfrac{995}{994}\) 

= \(\dfrac{1991\times1992\times1993\times1994\times995}{1991\times1992\times1993\times994\times995\times2}\)

= \(\dfrac{1994}{994\times2}\)

= \(\dfrac{997}{994}\)

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2011}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2011}\)

=>x+1=2011

=>x=2010

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\Rightarrow x+1=2011\Rightarrow x=2010\)

Vậy x=2010