1/3 = 2/6 = 2/(2x3) = 2/2 - 2/3
1/6 = 2/12 = 2/(3x4) = 2/3 - 2/4
...
2/x(x + 1) = 2/x - 2/(x +1)
Do đó: 
1/3 + 1/6 + ... + 2/x(x+1) = 2/2 - 2/3 + 2/3 - 2/4 + ... +2/x - 2/(x + 1) = 2/2 - 2/(x+1)
suy ra 1 - 2/(x + 1) = 2013/2014

x= 4027

1/3 = 2/6 = 2/(2x3) = 2/2 - 2/3 1/6 = 2/12 = 2/(3x4) = 2/3 - 2/4 ... 2/x(x + 1) = 2/x - 2/(x +1) Do đó: 1/3 + 1/6 + ... + 2/x(x+1) = 2/2 - 2/3 + 2/3 - 2/4 + ... +2/x - 2/(x + 1) = 2/2 - 2/(x+1) suy ra 1 - 2/(x + 1) = 2013/2014 x= 4027

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

=> \(\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2011}{2013}\)

=> \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)

=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)

=> x+1 = 2013 => x = 2012

kết quả là 99999999999

1/3+1/6+1/10+...........+1/1300

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{2}-\frac{1}{3}+...-\frac{1}{x+1}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2x-2}{2x+2}=\frac{2}{2013}\left(\text{vô nghiệm}\right);\frac{1}{3}>\frac{2}{2013}\text{ do đó vô nghiệm}\left(\text{ngắn hơn :))}\right)\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x+\left(x+1\right)}=\frac{2}{2013}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{2013}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2}{2013}\)

\(\Rightarrow\frac{2x-2}{2x+2}=\frac{2}{2013}\)

\(\Rightarrow\frac{x-1}{x+1}=\frac{2}{2013}\left(vl\right)\)

=> Bt trên có x vô nghiệm

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{2013}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)

\(\Rightarrow\frac{x-1}{2x+2}=\frac{1}{2013}\)

\(\Rightarrow2013\left(x-1\right)=2x+2\)

\(\Rightarrow2013x-2013=2x+2\)

\(\Rightarrow2011x=2015\)

\(\Rightarrow x=\frac{2015}{2011}\) 

C=(2x-1)(x-1)(2x^2-3x-1)+2017

=(2x^2-3x+1)(2x^2-3x-1)+2017

=(2x^2-3x)^2-1+2017

=(2x^2-3x)^2+2016>=2016

Dấu = xảy ra khi 2x^2-3x=0

=>x=0 hoặc x=3/2

D=(x-1)(x-6)(x-3)(x-4)+10

=(x^2-7x+6)(x^2-7x+12)+10

=(x^2-7x)^2+18*(x^2-7x)+72+10

=(x^2-7x+9)^2+1>=1

Dấu = xảy ra khi x^2-7x+9=0

=>\(x=\dfrac{7\pm\sqrt{13}}{2}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2015}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)

tự làm tiếp nhé mk ăn cơm đã

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

=> \(2.\frac{1}{2}-2.\frac{1}{x+1}=\frac{2011}{2013}\)

=> \(1-\frac{2}{x+1}=\frac{2011}{2013}\)

=> \(\frac{2}{x+1}=1-\frac{2011}{2013}=\frac{2}{2013}\)

=> x + 1 = 2013

=> x = 2013 - 1 = 2012