Chọn B

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

\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)

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

\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)

\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)

\(=x^{2n}-2x^n+x^n-2-x^{2n}+x^n+2018\)

\(=\left(x^{2n}-x^{2n}\right)+\left(-2x^n+x^n+x^2\right)+\left(-2+2018\right)\)

\(=2016\)

Vậy BT trên k phụ thuộc vào biến

\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)

\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)

\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)

Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:

\(N=\dfrac{\left(y-1\right)\left(y+1\right)+1}{y}\)

\(=\dfrac{y^2-1+1}{y}\)

\(=\dfrac{y^2}{y}\)

\(=y\)

\(=x^2+7x+11\)

Vậy \(N=x^2+7x+11\).

\(\text{#}Toru\)

Giải:

\(\left(x^n+1\right)\left(x^n-2\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2018\)

\(=x^{2n}+x^n-2x^n-2-\left(x^{n-3}.x^{n+3}\right)+x^{n-3}.x^3+2018\)

\(=x^{2n}-x^n-2-x^{n-3+n+3}+x^{n-3+3}+2018\)

\(=x^{2n}-x^n-2-x^{2n}+x^n+2018\)

\(=-2+2018\)

\(=2016\)

Vậy ...

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

(x² + 1)(x - 3) - (x - 3)(x² - 1)
= [x² + 1 - (x² - 1)](x - 3)

= (x² + 1 - x² + 1)(x - 3)

= 2(x - 3)

`=(x-3)[x^2+1-(x^2-1)]`

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

`=2(x-3)`