Với $x,y$ thỏa mãn hệ phương trình $\left\{ \begin{array}{l}A_x^2 + C_y^3 = 22\\A_y^3 + C_x^2 = 66\end{array} \right.\,\,\,\left( {x,y \in N} \right)$ thì $x - y$ bằng?

ĐK: $x \ge 2,y \ge 3,x,y \in N$

Ta có: $C_x^2 = \dfrac{1}{{2!}}A_x^2 = \dfrac{1}{2}A_x^2\,;\,C_y^3 = \dfrac{1}{{3!}}A_y^3 = \dfrac{1}{6}A_y^3$

Đặt $A_x^2 = a\,;\,A_y^3 = b$ ta có:

$hpt \Leftrightarrow \left\{ \begin{array}{l}a + \dfrac{b}{6} = 22\\b + \dfrac{a}{2} = 66\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 12\\b = 60\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A_x^2 = 12\,\,\left( 1 \right)\\A_y^3 = 60\,\,\left( 2 \right)\end{array} \right.$

Giải (1):

$A_x^2 = 12 \Leftrightarrow \dfrac{{x!}}{{\left( {x - 2} \right)!}} = 12 \Leftrightarrow x\left( {x - 1} \right) = 12 \Leftrightarrow {x^2} - x - 12 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 4\,\,\,\,\,\left( {tm} \right)\\x =  - 3\,\,\left( {ktm} \right)\end{array} \right.$

Giải (2): 

$\begin{array}{l}A_y^3 = 60 \Leftrightarrow \dfrac{{y!}}{{\left( {y - 3} \right)!}} = 60\\ \Leftrightarrow y\left( {y - 1} \right)\left( {y - 2} \right) = 60\\ \Leftrightarrow {y^3} - 3{y^2} + 2y - 60 = 0\\ \Leftrightarrow y = 5\,\,\left( {tm} \right)\end{array}$

Vậy $x - y = 4 - 5 =  - 1$