Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(n^2+2n-4=n^2+2n-15+11=\left(n-3\right)\left(n-5\right)+11\)
Để \(n^2+2n-4⋮11\Leftrightarrow\left(n-3\right)\left(n+5\right)⋮11\Leftrightarrow\left[{}\begin{matrix}n-3⋮11\\n+5⋮11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=BS11+3\\n=BS11-5\end{matrix}\right.\)
c,\(\dfrac{n^3-n^2+2n+7}{n^2+1}=\dfrac{n^3+n-n^2-1+n+8}{n^2+1}=\dfrac{n\left(n^2+1\right)-\left(n^2+1\right)+n+8}{n^2+1}=n-1+\dfrac{n+8}{n^2+1}\)
Để \(n^3-n^2+2n+7⋮n^2+1\Leftrightarrow n+8⋮n^2+1\)
\(\Rightarrow\left(n+8\right)\left(n-8\right)⋮n^2+1\Rightarrow n^2-64⋮n^2+1\)
\(\Rightarrow n^2+1-65⋮n^2+1\Rightarrow65⋮n^2+1\)
\(\Rightarrow n^2+1\inƯ\left(65\right)=\left\{\pm1;\pm5;\pm13;\pm65\right\}\)
Mà \(n^2+1\ge1\Rightarrow n^2+1\in\left\{1;5;13;65\right\}\)
\(\Rightarrow n\in\left\{0;\pm2;\sqrt{12};\pm8\right\}\)
\(a,\left(2x-3\right)n-2n\left(n+2\right)\)
\(=n\left(2x-3-2n-4\right)\)
\(=-7n\)
Vì \(-7⋮7\Rightarrow-7n⋮7\) => ĐPCM
\(b,n\left(2n-3\right)-2n\left(n+1\right)\)
\(=n\left(2n-3-2n-2\right)\)
\(=-5n⋮5\) (ĐPCM)
Rút gọn
\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)
\(=-76\)
\(b,\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)
\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)
\(=9\)
\(c,3x^2\left(x^2+2\right)+4x\left(x^2-1\right)-\left(x^2+2x+3\right)\left(3x^2-2x+1\right)\)
\(=3x^4+6x^2+4x^3-4x-3x^4+2x^3-x^2-6x^3+4x^2-2x-9x^2+6x-3\)
= -3