30.40=1200

Another way:

We have: \(\overline{x_nx_{n-1}...x_3x_2x_1}=100.\overline{x_nx_{n-1}...x_3}+\overline{x_2x_1}\)

Since \(100⋮4\)=>\(100.\overline{x_nx_{n-1}...x_3}⋮4\)

\(\overline{x_nx_{n-1}...x_3x_2x_1}⋮4\)=>\(\overline{x_2x_1}⋮4\)

In other word: a number is divisibility by 4 when the last three digits on the right-hand is divisibility by 4

We have: \(\dfrac{2n+1}{\left(n^2+n\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

So: \(M=\dfrac{2.1+1}{\left(1^2+1\right)^2}+\dfrac{2.2+1}{\left(2^2+2\right)^2}+...+\dfrac{2.2015+1}{\left(2015^2+2015\right)^2}\)

            \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)

            \(=1-\dfrac{1}{2016^2}< 1\)

Condition: \(x\ne5\)

We have: \(\dfrac{x^2-5x}{x-5}=5\Leftrightarrow\dfrac{x\left(x-5\right)}{x-5}=5\Leftrightarrow x=5\)

So the experiment of the above equation is \(S=\left\{\phi\right\}\)

We have: \(A=x^4+6x^{^3}+7x^2-6x+1\)

                    \(=\left(x^4+6x^3+9x^2\right)-\left(2x^2+6x\right)+1\)

                    \(=\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1\)

                    \(=\left(x^2+3x-1\right)^2\)

Ta có: \(\dfrac{2n+1}{n^2\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

Do đó \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

       \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

       \(=1-\dfrac{1}{\left(n+1\right)^2}< 1\)

Vậy \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\forall n\in\)N*