腾讯课堂目标2017初中数学联赛集训队作业题解答-9

Posted 2020-08-22 赵胤数学课

 

课程链接：目标2017初中数学联赛集训队-1（赵胤授课）

 

 

1. 已知 $a$, $b$, $c$ 为整数, 且 $a+b = 2006$, $c - a = 2005$. 若 $a < b$, 则 $a + b + c$ 之最大值是多少?

解答:

$\because a, b\in\mathbf{Z}$, $a + b = 2006$, $\therefore a \le 1002$, 因此由 $$\begin{cases}a + b = 2006\\ c - a = 2005\end{cases}$$ $$\Rightarrow a + b + c = 2006 + a + 2005 = a + 4011 \le 1002 + 4011 = 5013.$$ 即 $a + b + c$ 之最大值为 $5013$.

 

 

2. 解方程: $$[2x] + [3x] = 8x - {7 \over 2}.$$ 解答: $$\begin{cases}2x - 1 < [2x] \le 2x\\ 3x - 1 < [3x] \le 3x \end{cases} \Rightarrow 5x - 2 < [2x] + [3x] \le 5x$$ $$\Rightarrow 5x - 2 < 8x - {7\over2} \le 5x \Rightarrow {1\over2} < x \le {7\over6}$$ $$\Rightarrow {1\over2} < 8x - {7\over2} \le {35\over6} = 5{5\over6} \Rightarrow 8x - {7\over2} = 1,\ 2,\ 3,\ 4,\ 5,$$ $$\Rightarrow x = {9\over16},\ {11\over16},\ {13\over16},\ {15\over16},\ {17\over16}.$$ 经检验, $x_1 = \displaystyle{13\over16}$, $x_2 = \displaystyle{17\over16}$ 是原方程的解.

 

 

3. 已知 $x = \displaystyle{b\over a}$, ($a$, $b$ 互素), 且 $a \le 8$, $\sqrt2 - 1 < x < \sqrt3 - 1$. 求所有满足条件的 $x$.

解答: $$\sqrt2 - 1 < x < \sqrt3 - 1 \Rightarrow \sqrt2 - 1 < {b \over a} < \sqrt3 - 1$$ $$\Rightarrow (\sqrt2 - 1)a < b < (\sqrt3 - 1)a$$ 分别将 $a = 1, 2, \cdots, 8$ 代入验证, 可得所有满足题意的 $x$ 值如下: $${1\over2},\ {2\over3},\ {3\over5},\ {3\over7},\ {4\over7},\ {5\over7},\ {5\over8}.$$

 

4. $n$ ($n > 1$)个整数 $a_1$, $a_2$, $\cdots$, $a_n$ 满足 $a_1 + a_2 + \cdots + a_n = a_1 a_2 \cdots a_n = 2007$, 求 $n$ 的最小值.

解答:

易知 $a_i$ 均为奇数, 且 $n$ 为奇数.

若 $n = 3$, 即 $a_1 + a_2 + a_3 = a_1a_2a_3 = 2007$, 不妨设 $a_1 \ge a_2 \ge a_3$, 则 $$a_1 \ge {2007 \over 3} = 669,\ a_2a_3 \le 3.$$ 若 $a_1 = 669$, $a_2a_3 = 3$, 则 $$a_2 + a_3 \le 4 \Rightarrow a_1 + a_2 + a_3 \le 673 < 2007.$$ 若 $a_1 > 669$, 则 $a_1 = 2007$, $a_2a_3 = 1$ 且 $a_2 + a_3 = 0$, 矛盾.

由此可知 $n \ge 5$. 当 $a_1 = 2007$, $a_2 = a_3 = 1$, $a_4 = a_5 = -1$ 时满足题意.

因此 $n$ 的最小值为 $5$.

 

 

5. 求三位数与它的数字和之比的最小值.

解答:

设三位数为 $N = \overline{xyz}$, 则 $$\lambda = {100x + 10y + z \over x+ y+z} = 1 + {99x + 9y \over x + y + z}\ge 1 + {99x + 9y \over x + y + 9}$$ $$= 1 + {9(11x + y) \over x + y + 9} = 1 + 9\left(1 + {10x - 9 \over x + y + 9}\right)$$ $$\ge 1 + 9\left(1 + {10x - 9 \over x + 9 + 9}\right) = 1 + 9\left(1 + {10x - 9 \over x + 18}\right)$$ $$= 10 + 9\cdot{10x - 9 \over x + 18} = 10 + 9\cdot \left(10 - {189 \over x + 18}\right)$$ $$= 100 - {1701 \over x + 18} \ge 100 - {1701 \over 1+18} = {199 \over 19}.$$ 因此当三位数为 $199$ 时可取到最小比值 $\lambda = \displaystyle10{9 \over 19}$.

 

 

6. 若 $a > 0$, $b > 0$, 且 $h = \min\left(a, \displaystyle{b \over a^2 + b^2}\right)$, 求 $h$ 之最大值.

解答:

分类讨论即可.

若 $h = \min\left(a, \displaystyle{b \over a^2 + b^2}\right) = a$, 则 $$h^2 = a^2 \le {ab \over a^2 + b^2} \le {ab \over 2ab} = {1\over2}$$ $$\Rightarrow h \le {\sqrt2 \over 2}.$$ 若 $h = \min\left(a, \displaystyle{b \over a^2 + b^2}\right) = \displaystyle{b \over a^2 + b^2}$, 则 $$h^2 = \left({b \over a^2 + b^2}\right)^2 \le {ab \over a^2 + b^2} \le {ab \over 2ab} = {1\over2}$$ $$\Rightarrow h \le {\sqrt2 \over 2}.$$ 综上, $h$ 之最大值为 $\displaystyle{\sqrt2 \over 2}$.