Tìm điều kiện để phép chia là phép chia hết:

a) xⁿyⁿ⁺¹ : x²y⁵;

b) (13x⁴y³ – 5x³y³ + 6x²y²) : 5xⁿyⁿ.

a) Ta có: xⁿyⁿ⁺¹ ⋮ x²y⁵

$\Rightarrow \left\{ {\begin{array}{*{20}{c}}{n \ge 2}\\{n + 1 \ge 5}\end{array}} \right.$$\Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{n \ge 2}\\{n \ge 4}\end{array}} \right.$⇔n ≥ 4

b) Ta có: (13x⁴y³ – 5x³y³ + 6x²y²) ⋮ 5xⁿyⁿ

$\Rightarrow \left\{ {\begin{array}{*{20}{c}}{13{x^4}{y^3} \vdots 5{x^n}{y^n}}\\{5{x^3}{y^3} \vdots 5{x^n}{y^n}}\\{6{x^2}{y^2} \vdots 5{x^n}{y^n}}\end{array}} \right.$

$\Rightarrow \left\{ {\begin{array}{*{20}{c}}{n \le 4}\\{n \le 3}\\{n \le 2}\end{array}} \right.$⇔n ≤ 2