Question

Giải với

Answer 1

a) 

\(\dfrac{A}{B}=\dfrac{-13x^{17}y^{2n-3}+22x^{16}y^7}{-7x^{3n+1}y^6}=\dfrac{13}{7}x^{17-\left(3n+1\right)}y^{2n-3-6}-\dfrac{22}{7}x^{16-\left(3n+1\right)}y^{7-6}\\ =\dfrac{13}{7}x^{16-3n}y^{2n-9}-\dfrac{22}{7}x^{15-3n}y\)

Để A chia hết cho b thì: \(\left\{{}\begin{matrix}16-3n\ge0\\2n-9\ge0\\15-3n\ge0\end{matrix}\right.\Leftrightarrow\dfrac{9}{2}\le n\le5\Rightarrow n=5\) (vì n là STN) 

b) 

\(\dfrac{A}{B}=\dfrac{20x^8y^{2n}-10x^4y^{3n}+15x^5y^6}{3x^{2n}y^{n+1}}\\ =\dfrac{20}{3}x^{8-2n}y^{2n-\left(n+1\right)}-\dfrac{10}{3}x^{4-2n}y^{3n-\left(n+1\right)}+5x^{5-2n}y^{6-\left(n+1\right)}\\ =\dfrac{20}{3}x^{8-2n}y^{n-1}-\dfrac{10}{3}x^{4-2n}y^{2n-1}+5^{5-2n}y^{5-n}\)

Để A chia hết cho B thì: 

\(8-2n\ge0;n-1\ge0;4-2n\ge0;2n-1\ge0;5-2n\ge0;5-n\ge0\) 

\(\Rightarrow1\le n\le2\Rightarrow n\in\left\{1;2\right\}\)