Question

Tìm số tự nhiên n để phép chia sau là phép chia hết:

\(\left(x^{12}y^{n+2}-5x^{10}y^{n-1}+3x^9y^{n-2}\right):2x^{n-2}y^3\)

Answer 1

\(=x^2y^{n+2}:2x^{n-2}y^3-5x^{10}.y^{n-1}:2x^{n-2}.y^3+3x^9y^{n-2}:2x^{n-2}y^3\)
\(=\dfrac{1}{2}x^{4-n}y^{n-1}-\dfrac{5}{2}x^{12-n}y^{n-4}+\dfrac{3}{2}x^{11-n}y^{n-5}\)
Ta có \(4-n\ge0< =>x\le4\)
\(n-1\ge0< =>n\ge1\)          
\(12-n\ge0< =>n\le12\)
\(n-4\ge0< =>n\ge4\)
\(11-n\ge0< =>n\ge11\)
\(n-5\ge0< =>n\ge5\)
\(=>5\le n\le11=>n\in\left\{5;6;7;8;9;10;11\right\}\)