1. 已知 $ m $，$ n $ 均为正整数且满足 $ mn - 3m - 2n - 24 = 0 $，则 $ m + n $ 的最大值是(
A.16
B.22
C.34
D.36
D
)A.16
B.22
C.34
D.36
答案：1. D 点拨:将方程左边变形,得 $ mn - 3m - 2n + 6 - 30 = m(n - 3) - 2(n - 3) - 30 $,
$\therefore (m - 2)(n - 3) = 30$.
$\because m,n$ 均为正整数,
$\therefore \{\begin{array}{l} m - 2 = 1,\\ n - 3 = 30\end{array} $ 或 $\{\begin{array}{l} m - 2 = 2,\\ n - 3 = 15\end{array} $ 或 $\{\begin{array}{l} m - 2 = 3,\\ n - 3 = 10\end{array} $
或 $\{\begin{array}{l} m - 2 = 5,\\ n - 3 = 6\end{array} $ 或 $\{\begin{array}{l} m - 2 = 30,\\ n - 3 = 1\end{array} $ 或 $\{\begin{array}{l} m - 2 = 15,\\ n - 3 = 2\end{array} $
或 $\{\begin{array}{l} m - 2 = 10,\\ n - 3 = 3\end{array} $ 或 $\{\begin{array}{l} m - 2 = 6,\\ n - 3 = 5,\end{array} $
解得 $\{\begin{array}{l} m = 3,\\ n = 33\end{array} $ 或 $\{\begin{array}{l} m = 4,\\ n = 18\end{array} $ 或 $\{\begin{array}{l} m = 5,\\ n = 13\end{array} $ 或 $\{\begin{array}{l} m = 7,\\ n = 9\end{array} $
或 $\{\begin{array}{l} m = 32,\\ n = 4\end{array} $ 或 $\{\begin{array}{l} m = 17,\\ n = 5\end{array} $ 或 $\{\begin{array}{l} m = 12,\\ n = 6\end{array} $ 或 $\{\begin{array}{l} m = 8,\\ n = 8,\end{array} $
$\therefore m + n = 36$ 或 $ 22 $ 或 $ 18 $ 或 $ 16 $,
$\therefore m + n$ 的最大值是 $ 36 $.
$\therefore (m - 2)(n - 3) = 30$.
$\because m,n$ 均为正整数,
$\therefore \{\begin{array}{l} m - 2 = 1,\\ n - 3 = 30\end{array} $ 或 $\{\begin{array}{l} m - 2 = 2,\\ n - 3 = 15\end{array} $ 或 $\{\begin{array}{l} m - 2 = 3,\\ n - 3 = 10\end{array} $
或 $\{\begin{array}{l} m - 2 = 5,\\ n - 3 = 6\end{array} $ 或 $\{\begin{array}{l} m - 2 = 30,\\ n - 3 = 1\end{array} $ 或 $\{\begin{array}{l} m - 2 = 15,\\ n - 3 = 2\end{array} $
或 $\{\begin{array}{l} m - 2 = 10,\\ n - 3 = 3\end{array} $ 或 $\{\begin{array}{l} m - 2 = 6,\\ n - 3 = 5,\end{array} $
解得 $\{\begin{array}{l} m = 3,\\ n = 33\end{array} $ 或 $\{\begin{array}{l} m = 4,\\ n = 18\end{array} $ 或 $\{\begin{array}{l} m = 5,\\ n = 13\end{array} $ 或 $\{\begin{array}{l} m = 7,\\ n = 9\end{array} $
或 $\{\begin{array}{l} m = 32,\\ n = 4\end{array} $ 或 $\{\begin{array}{l} m = 17,\\ n = 5\end{array} $ 或 $\{\begin{array}{l} m = 12,\\ n = 6\end{array} $ 或 $\{\begin{array}{l} m = 8,\\ n = 8,\end{array} $
$\therefore m + n = 36$ 或 $ 22 $ 或 $ 18 $ 或 $ 16 $,
$\therefore m + n$ 的最大值是 $ 36 $.
2. 用分组分解法或十字相乘法分解因式：
(1) $ ax + bx + 3a + 3b $；
(2) $ 4a^{2} - b^{2} + 6a - 3b $；
(3) $ x^{2} - 4y^{2} + 12yz - 9z^{2} $；
(4) $ 2x^{2} + 5xy + 3y^{2} $；
(5) $ 5a^{2}b^{2} + 23ab - 10 $；
(6) $ 4x^{4} - 13x^{2}y^{2} + 9y^{4} $；
(7) $ 4m^{2} - 4m - 4n^{2} + 1 $；
(8) $ x^{3} + 6x^{2}y - 27xy^{2} $。
(1) $ ax + bx + 3a + 3b $；
(2) $ 4a^{2} - b^{2} + 6a - 3b $；
(3) $ x^{2} - 4y^{2} + 12yz - 9z^{2} $；
(4) $ 2x^{2} + 5xy + 3y^{2} $；
(5) $ 5a^{2}b^{2} + 23ab - 10 $；
(6) $ 4x^{4} - 13x^{2}y^{2} + 9y^{4} $；
(7) $ 4m^{2} - 4m - 4n^{2} + 1 $；
(8) $ x^{3} + 6x^{2}y - 27xy^{2} $。
答案：2. 解:(1) 原式 $ = (ax + bx) + (3a + 3b) = x(a + b) + 3(a + b) = (a + b)(x + 3) $.
(2) 原式 $ = (2a + b)(2a - b) + 3(2a - b) = (2a - b) · (2a + b + 3) $.
(3) 原式 $ = x^{2} - (4y^{2} - 12yz + 9z^{2}) = x^{2} - (2y - 3z)^{2} = (x + 2y - 3z)(x - 2y + 3z) $.
(4) 原式 $ = (x + y)(2x + 3y) $.
(5) 原式 $ = (ab + 5)(5ab - 2) $.
(6) 原式 $ = (x^{2} - y^{2})(4x^{2} - 9y^{2}) = (x + y)(x - y)(2x + 3y)(2x - 3y) $.
(7) 原式 $ = (4m^{2} - 4m + 1) - 4n^{2} = (2m - 1)^{2} - (2n)^{2} = (2m - 1 + 2n)(2m - 1 - 2n) $.
(8) 原式 $ = x(x^{2} + 6xy - 27y^{2}) = x(x - 3y)(x + 9y) $.
(2) 原式 $ = (2a + b)(2a - b) + 3(2a - b) = (2a - b) · (2a + b + 3) $.
(3) 原式 $ = x^{2} - (4y^{2} - 12yz + 9z^{2}) = x^{2} - (2y - 3z)^{2} = (x + 2y - 3z)(x - 2y + 3z) $.
(4) 原式 $ = (x + y)(2x + 3y) $.
(5) 原式 $ = (ab + 5)(5ab - 2) $.
(6) 原式 $ = (x^{2} - y^{2})(4x^{2} - 9y^{2}) = (x + y)(x - y)(2x + 3y)(2x - 3y) $.
(7) 原式 $ = (4m^{2} - 4m + 1) - 4n^{2} = (2m - 1)^{2} - (2n)^{2} = (2m - 1 + 2n)(2m - 1 - 2n) $.
(8) 原式 $ = x(x^{2} + 6xy - 27y^{2}) = x(x - 3y)(x + 9y) $.