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24. (6分)若$M = (x - 2)(x - 7)$,$N = (x - 6)(x - 3)$,则$M$与$N$谁大?
解:$\because M - N = (x - 2)(x - 7) - (x - 6)(x - 3)$
$ = x^{2} - 9x + 14 - (x^{2} - 9x + 18) = x^{2} - 9x + 14 - x^{2} + 9x - 18 = -4 ＜ 0$,
$\therefore M - N ＜ 0$,
$\therefore M ＜ N$.
解:$\because M - N = (x - 2)(x - 7) - (x - 6)(x - 3)$
$ = x^{2} - 9x + 14 - (x^{2} - 9x + 18) = x^{2} - 9x + 14 - x^{2} + 9x - 18 = -4 ＜ 0$,
$\therefore M - N ＜ 0$,
$\therefore M ＜ N$.
答案：
$\because M - N=(x - 2)(x - 7)-(x - 6)(x - 3)$
$=x^{2}-9x + 14-(x^{2}-9x + 18)=x^{2}-9x + 14 - x^{2}+9x - 18=-4＜0$，
$\therefore M - N＜0$，
$\therefore M＜N$.
$=x^{2}-9x + 14-(x^{2}-9x + 18)=x^{2}-9x + 14 - x^{2}+9x - 18=-4＜0$，
$\therefore M - N＜0$，
$\therefore M＜N$.
25. (8分)中考新考法 解题方法型阅读理解题 阅读材料:若$m^{2} - 2mn + 2n^{2} - 8n + 16 = 0$,求$m$,$n$的值.
解:$\because m^{2} - 2mn + 2n^{2} - 8n + 16 = 0$,
$\therefore (m^{2} - 2mn + n^{2}) + (n^{2} - 8n + 16) = 0$,
$\therefore (m - n)^{2} + (n - 4)^{2} = 0$.
$\because (m - n)^{2} \geq 0$,$(n - 4)^{2} \geq 0$,
$\therefore (m - n)^{2} = 0$,$(n - 4)^{2} = 0$,
$\therefore n = 4$,$m = 4$.
根据你的观察,探究下面的问题:
(1)比较大小:$x^{2} + 1 \geq 2x$;$x^{2} - 6x \geq -9$.
(2)已知:$x^{2} + 2xy + 2y^{2} + 2y + 1 = 0$,求$2x + 3y$的值.
(3)已知:$a - c = 6$,$ac + b^{2} - 8b + 25 = 0$,求$a + 2b + 3c$的值.
解:(2)$\because x^{2} + 2xy + 2y^{2} + 2y + 1 = 0$,
$\therefore x^{2} + 2xy + y^{2} + y^{2} + 2y + 1 = 0$,
$\therefore (x + y)^{2} + (y + 1)^{2} = 0$,
$\therefore x + y = 0$,$y + 1 = 0$,解得$x = 1$,$y = -1$,
$\therefore 2x + 3y = 2 \times 1 + 3 \times (-1) = -1$.
(3)$\because a - c = 6$,
$\therefore a = c + 6$.
$\because ac + b^{2} - 8b + 25 = 0$,
$\therefore (c + 6)c + b^{2} - 8b + 25 = 0$,
$\therefore c^{2} + 6c + 9 + b^{2} - 8b + 16 = 0$,
$\therefore (c + 3)^{2} + (b - 4)^{2} = 0$,
$\therefore c + 3 = 0$,$b - 4 = 0$,
$\therefore c = -3$,$b = 4$,$a = 3$,
$\therefore a + 2b + 3c = 3 + 2 \times 4 + 3 \times (-3) = 2$.
解:$\because m^{2} - 2mn + 2n^{2} - 8n + 16 = 0$,
$\therefore (m^{2} - 2mn + n^{2}) + (n^{2} - 8n + 16) = 0$,
$\therefore (m - n)^{2} + (n - 4)^{2} = 0$.
$\because (m - n)^{2} \geq 0$,$(n - 4)^{2} \geq 0$,
$\therefore (m - n)^{2} = 0$,$(n - 4)^{2} = 0$,
$\therefore n = 4$,$m = 4$.
根据你的观察,探究下面的问题:
(1)比较大小:$x^{2} + 1 \geq 2x$;$x^{2} - 6x \geq -9$.
(2)已知:$x^{2} + 2xy + 2y^{2} + 2y + 1 = 0$,求$2x + 3y$的值.
(3)已知:$a - c = 6$,$ac + b^{2} - 8b + 25 = 0$,求$a + 2b + 3c$的值.
解:(2)$\because x^{2} + 2xy + 2y^{2} + 2y + 1 = 0$,
$\therefore x^{2} + 2xy + y^{2} + y^{2} + 2y + 1 = 0$,
$\therefore (x + y)^{2} + (y + 1)^{2} = 0$,
$\therefore x + y = 0$,$y + 1 = 0$,解得$x = 1$,$y = -1$,
$\therefore 2x + 3y = 2 \times 1 + 3 \times (-1) = -1$.
(3)$\because a - c = 6$,
$\therefore a = c + 6$.
$\because ac + b^{2} - 8b + 25 = 0$,
$\therefore (c + 6)c + b^{2} - 8b + 25 = 0$,
$\therefore c^{2} + 6c + 9 + b^{2} - 8b + 16 = 0$,
$\therefore (c + 3)^{2} + (b - 4)^{2} = 0$,
$\therefore c + 3 = 0$,$b - 4 = 0$,
$\therefore c = -3$,$b = 4$,$a = 3$,
$\therefore a + 2b + 3c = 3 + 2 \times 4 + 3 \times (-3) = 2$.
答案：
(1)$\geqslant$ $\geqslant$ [解析]$\because x^{2}+1 - 2x=(x - 1)^{2}\geqslant0$，
$\therefore x^{2}+1\geqslant2x$.
$\because x^{2}-6x-(-9)=x^{2}-6x + 9=(x - 3)^{2}\geqslant0$，
$\therefore x^{2}-6x\geqslant - 9$.
(2)$\because x^{2}+2xy + 2y^{2}+2y + 1 = 0$，
$\therefore x^{2}+2xy + y^{2}+y^{2}+2y + 1 = 0$，
$\therefore(x + y)^{2}+(y + 1)^{2}=0$，
$\therefore x + y = 0$，$y + 1 = 0$，解得 $x = 1$，$y = - 1$，
$\therefore2x + 3y = 2\times1+3\times(-1)=-1$.
(3)$\because a - c = 6$，
$\therefore a = c + 6$.
$\because ac + b^{2}-8b + 25 = 0$，
$\therefore(c + 6)c + b^{2}-8b + 25 = 0$，
$\therefore c^{2}+6c + 9 + b^{2}-8b + 16 = 0$，
$\therefore(c + 3)^{2}+(b - 4)^{2}=0$，
$\therefore c + 3 = 0$，$b - 4 = 0$，
$\therefore c = - 3$，$b = 4$，$a = 3$，
$\therefore a + 2b + 3c = 3+2\times4+3\times(-3)=2$.
思路引导 本题考查了配方法的应用，任意一个数的偶次方都是非负数，当几个数或式的偶次方相加和为 0 时，则其中的每一项都必须等于 0，熟练掌握完全平方公式是解本题的关键.
(1)$\geqslant$ $\geqslant$ [解析]$\because x^{2}+1 - 2x=(x - 1)^{2}\geqslant0$，
$\therefore x^{2}+1\geqslant2x$.
$\because x^{2}-6x-(-9)=x^{2}-6x + 9=(x - 3)^{2}\geqslant0$，
$\therefore x^{2}-6x\geqslant - 9$.
(2)$\because x^{2}+2xy + 2y^{2}+2y + 1 = 0$，
$\therefore x^{2}+2xy + y^{2}+y^{2}+2y + 1 = 0$，
$\therefore(x + y)^{2}+(y + 1)^{2}=0$，
$\therefore x + y = 0$，$y + 1 = 0$，解得 $x = 1$，$y = - 1$，
$\therefore2x + 3y = 2\times1+3\times(-1)=-1$.
(3)$\because a - c = 6$，
$\therefore a = c + 6$.
$\because ac + b^{2}-8b + 25 = 0$，
$\therefore(c + 6)c + b^{2}-8b + 25 = 0$，
$\therefore c^{2}+6c + 9 + b^{2}-8b + 16 = 0$，
$\therefore(c + 3)^{2}+(b - 4)^{2}=0$，
$\therefore c + 3 = 0$，$b - 4 = 0$，
$\therefore c = - 3$，$b = 4$，$a = 3$，
$\therefore a + 2b + 3c = 3+2\times4+3\times(-3)=2$.
思路引导 本题考查了配方法的应用，任意一个数的偶次方都是非负数，当几个数或式的偶次方相加和为 0 时，则其中的每一项都必须等于 0，熟练掌握完全平方公式是解本题的关键.
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