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. $$|x|y + x^3 = 0Rightarrow y = -|x|x $$ 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 $$-x + x^2 = 3 Rightarrow x^2 - x - 3 = 0 Rightarrow egin{cases}x = {1 - sqrt{13} over 2}\ y = {7 - sqrt{13} over 2} end{cases} Rightarrow x + y = 4 - sqrt{13}.$$ 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 $$(b - a_1)(b - a_2)(b - a_3)(b - a_4)(b - a_5) = 2009 = 7^2 imes 41$$ $$Rightarrow (b - a_1)(b - a_2)(b - a_3)(b - a_4)(b - a_5) = -1 imes 1 imes 7 imes (-7) imes 41$$ $$Rightarrow (b - a_1) + (b - a_2) + (b - a_3) + (b - a_4) + (b - a_5) = 41Rightarrow 5b = 41 + 9$$ 因此 $b = 10$.
3. 已知实数 $a e b$, 且满足 $(a + 1)^2 = 3 - 3(a+1)$, $3(b + 1) = 3 - (b + 1)^2$. 则 $displaystyle bsqrt{bover a} + asqrt{a over b} = ?$
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 $$egin{cases}a + b = -5\ ab = 1 end{cases}Rightarrow a < 0, b < 0.$$ $$Rightarrow bsqrt{bover a} + asqrt{a over b} = {-(a^2 + b^2) over sqrt{ab}} = {-23 over 1} = -23.$$
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 $sqrt{2x^2 - 5x + 2} = a$, $sqrt{x^2 - 7x + 6} = b$, $sqrt{2x^2 - 3x + 1} = c$, $sqrt{x^2 - 9x + 7} = d$. $$egin{cases}a + b = c + d\ a^2 + b^2 = c^2 + d^2 end{cases} Rightarrow egin{cases}a - c = d - b\ a^2 - c^2 = d^2 - b^2end{cases}$$ 若 $a - c = d - b = 0$, 则 $x = 0.5$.
若 $a - c = d - b e 0$, 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: $$sqrt{10} = {x - 7 over sqrt{x - 3} + 2} + {x - 5 over sqrt{x - 4} + 1}$$ $$= {(x - 7)(sqrt{x - 3} - 2)over(sqrt{x-3} + 2)(sqrt{x - 3} - 2)} + {(x - 5)(sqrt{x - 4} - 1) over (sqrt{x - 4} + 1)(sqrt{x - 4} - 1)}$$ $$= sqrt{x - 3} - 2 + sqrt{x - 4} - 1$$ $$Rightarrow sqrt{x - 3} + sqrt{x - 4} = 3 + sqrt{10}$$ $$Rightarrow sqrt{x - 4} - sqrt{10} = 3 - sqrt{x - 3}$$ $$Rightarrow x + 6 - 2sqrt{10(x - 4)} = x + 6 - 6sqrt{x - 3}$$ $$Rightarrow 10(x - 4) = 9(x - 3) Rightarrow x = 13.$$ 经检验, $x = 13$ 是原方程的解.