腾讯课堂目标2017初中数学联赛集训队作业题解答-8

Posted 2020-08-21 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. 若 $x$, $y$ 是非零实数, 使得 $|x| + y = 3$ 和 $|x|y + x^3 = 0$, 则 $x + y = ?$

解答:

考虑去绝对值符号. $$|x|y + x^3 = 0\Rightarrow y = -|x|x $$ 若 $x \ge 0$, 则 $y = -x^2$, 代入得 $$x - x^2 = 3 \Rightarrow x^2 - x + 3 = 0 \Rightarrow \Delta = -11 < 0.$$ 若 $x < 0$, 则 $y = x^2$, 代入得 $$-x + x^2 = 3 \Rightarrow x^2 - x - 3 = 0 \Rightarrow \begin{cases}x = {1 - \sqrt{13} \over 2}\\ y = {7 - \sqrt{13} \over 2} \end{cases} \Rightarrow x + y = 4 - \sqrt{13}.$$ 综上, $x + y = 4 - \sqrt{13}$.

 

2. 已知 $a_1$, $a_2$, $a_3$, $a_4$, $a_5$ 是满足条件 $a_1 + a_2 + a_3 + a_4 + a_5 = 9$ 的五个不同的整数, 若 $b$ 是关于 $x$ 的方程 $(x - a_1)(x - a_2)(x - a_3)(x - a_4)(x - a_5) = 2009$ 的整数根, 则 $b=?$

解答:

由题意可得 $$(b - a_1)(b - a_2)(b - a_3)(b - a_4)(b - a_5) = 2009 = 7^2 \times 41$$ $$\Rightarrow (b - a_1)(b - a_2)(b - a_3)(b - a_4)(b - a_5) = -1 \times 1 \times 7 \times (-7) \times 41$$ $$\Rightarrow (b - a_1) + (b - a_2) + (b - a_3) + (b - a_4) + (b - a_5) = 41\Rightarrow 5b = 41 + 9$$ 因此 $b = 10$.

 

3. 已知实数 $a\ne b$, 且满足 $(a + 1)^2 = 3 - 3(a+1)$, $3(b + 1) = 3 - (b + 1)^2$. 则 $\displaystyle b\sqrt{b\over a} + a\sqrt{a \over b} = ?$

解答:

由题意可得 $a$, $b$ 是一元二次方程 $(x + 1)^2 + 3(x + 1) - 3 = 0 \Rightarrow x^2 +5x +1 = 0$ 的两根. 因此有 $$\begin{cases}a + b = -5\\ ab = 1 \end{cases}\Rightarrow a < 0,\ b < 0.$$ $$\Rightarrow b\sqrt{b\over a} + a\sqrt{a \over b} = {-(a^2 + b^2) \over \sqrt{ab}} = {-23 \over 1} = -23.$$

 

4. 解方程: $$\sqrt{2x^2 - 5x + 2} + \sqrt{x^2 - 7x + 6} = \sqrt{2x^2 - 3x + 1} + \sqrt{x^2 - 9x + 7}.$$ 解答:

换元法求解之.

令 $\sqrt{2x^2 - 5x + 2} = a$, $\sqrt{x^2 - 7x + 6} = b$, $\sqrt{2x^2 - 3x + 1} = c$, $\sqrt{x^2 - 9x + 7} = d$. $$\begin{cases}a + b = c + d\\ a^2 + b^2 = c^2 + d^2 \end{cases} \Rightarrow \begin{cases}a - c = d - b\\ a^2 - c^2 = d^2 - b^2\end{cases}$$ 若 $a - c = d - b = 0$, 则 $x = 0.5$.

若 $a - c = d - b \ne 0$, 则 $$a + c = d + b \Rightarrow a = d,\ b = c$$ $$\Rightarrow 2x^2 -5x + 2 = x^2 - 9x + 7 \Rightarrow x_1= -5, x_2 = 1.$$ 经检验, $x _ 1 = 0.5$, $x_2 = -5$ 是原方程的解.

 

5. 解方程: $${x - 7 \over \sqrt{x - 3} + 2} + {x - 5 \over \sqrt{x - 4} + 1} = \sqrt{10}.$$ 解答: $$\sqrt{10} = {x - 7 \over \sqrt{x - 3} + 2} + {x - 5 \over \sqrt{x - 4} + 1}$$ $$= {(x - 7)(\sqrt{x - 3} - 2)\over(\sqrt{x-3} + 2)(\sqrt{x - 3} - 2)} + {(x - 5)(\sqrt{x - 4} - 1) \over (\sqrt{x - 4} + 1)(\sqrt{x - 4} - 1)}$$ $$= \sqrt{x - 3} - 2 + \sqrt{x - 4} - 1$$ $$\Rightarrow \sqrt{x - 3} + \sqrt{x - 4} = 3 + \sqrt{10}$$ $$\Rightarrow \sqrt{x - 4} - \sqrt{10} = 3 - \sqrt{x - 3}$$ $$\Rightarrow x + 6 - 2\sqrt{10(x - 4)} = x + 6 - 6\sqrt{x - 3}$$ $$\Rightarrow 10(x - 4) = 9(x - 3) \Rightarrow x = 13.$$ 经检验, $x = 13$ 是原方程的解.