1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2005/2007

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2005/2007

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x*(x+1) = 2005/2007

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1) = 2005/2007

=> 2(1/2 - 1/x + 1) = 2005/2007

=> 1/2 - 1/x + 1 = 2005/4014

=> 1/x+1 = 1/2007

=> x + 1 = 2007

=> x = 2006

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2005}{2007}\) 

\(\rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\) 

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{2007}:2\) 

\(\rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{2007}:2\) \(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)

\(\Rightarrow x+1=2007\rightarrow x=2006\)

Vậy x = 2006.

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\(a,\left|x-2\right|< 4\)

\(\Rightarrow\left|x-2\right|\in\left(0;1;2;3\right)\)

Với \(\left|x-2\right|=0\Rightarrow x=2\)

\(Với\left|x-2\right|=1\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Với \(\left|x-2\right|=2\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

Với \(\left|x-2\right|=3\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

b,t tự

c,t tự

Ta có :\(\frac{1}{3}+\frac{1}{6}+..+\frac{2}{x\left(x+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)

          = 2 x \(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...\frac{1}{x\left(x+1\right)}\right)=2\times\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)\)

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

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

       Khi đó chỉ cần giải  2 x\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)

a) (x-1)³ = 125

     (x-1)³ = 5³

=> x-1 = 5

     x    = 5+1

     x    = 6

Vậy x = 6.

b) (2x+1)³ = 343

    (2x+1)³ = 7³

=> 2x + 1 = 7

     2x       = 7+1

     2x       = 8

     x         = 8 : 2

     x         = 4

Vậy x = 4

a)(x-1)^3=125

   (x-1)^3=5^3

    x-1=5

    x=5+1

    x=6

    vậy x=6

b)(2x+1)^3=343

   (2x-1)^3=7^3

   2x-1=7

   2x=7+1

   2x=8

   x=8:2

   x=4

 vậy x=4