猿辅导2017年春季初联训练营作业题解答-4: "一元二次方程－1"

Posted 2020-09-06 赵胤数学课

1. $x^3 - (\sqrt2 + 3)x^2 + (3\sqrt2 - 4)x + 4\sqrt2 = 0$

解答:

可参考例题2、例题9. $$f(-1) = 0 \Rightarrow f(x) = (x+1)\left[x^2 - (4+\sqrt2)x \right] = (x+1)(x-4)(x-\sqrt2)$$ $$\Rightarrow x_1 = -1,\ x_2 = 4,\ x_3 = \sqrt2.$$

 

 

2. $(6x + 7)^2(3x + 4)(x + 1) = 6$

解答:

可参考例题4.

令 $y = 6x + 7$, $$\Rightarrow y^2(y+1)(y-1) = 72 \Rightarrow y^4 - y^2 - 72 = 0 \Rightarrow (y^2 - 9)(y^2 + 8) = 0$$ $$\Rightarrow y = \pm3.$$ $$\Rightarrow x_1 = -\frac{2}{3},\ x_2 = -\frac{5}{3}.$$

 

 

3. $6x^4 + 5x^3 - 38x^2 + 5x + 6 = 0$

解答:

可参考例题8. $$6(x^2 + 1)^2 + 5x(x^2 + 1) - 50x^2 = 0$$ $$\Rightarrow \left[2(x^2+1) - 5x\right]\left[3(x^2+1) + 10x\right] = 0$$ $$\Rightarrow (2x^2 - 5x + 2)(3x^2 + 10x + 3) = 0$$ $$\Rightarrow x_1 = \frac{1}{2},\ x_2 = 2,\ x_3 = -\frac{1}{3},\ x_4 = -3.$$

 

 

4. $(x-2)(x-1)(x+3)(x+4) = 84$

解答:

可参考例题3. $$(x^2 + 2x - 8)(x^2 + 2x - 3) = 84$$ $$\Rightarrow (x^2 + 2x)^2 - 11(x^2 + 2x) - 60 = 0$$ $$\Rightarrow (x^2 + 2x - 15)(x^2 + 2x + 4) = 0$$ $$\Rightarrow x_1 = -5,\ x_2 = 3.$$

 

 

5. $x^4 + (x-4)^4 = 626$

解答:

可参考例题5.

令 $y = x-2$, $$\Rightarrow (y-2)^4 + (y+2)^4 = 626 \Rightarrow 2y^4 + 48y^2 - 594 = 0$$ $$\Rightarrow (y^2 + 33)(y^2 - 9) = 0 \Rightarrow y = \pm 3.$$ $$\Rightarrow x_1 = 5,\ x_2 = -1.$$

 

 

6. $4(2x^2 - 3x - 1)(x^2 - x + 2) - (3x^2 - 4x + 1)^2 = 0$

解答:

可参考例题6.

令 $m = 2x^2 - 3x - 1$, $n = x^2 - x + 2$, $$\Rightarrow 4mn - (m + n)^2 = 0$$ $$\Rightarrow m = n \Rightarrow 2x^2 - 3x - 1 = x^2 - x + 2$$ $$\Rightarrow x^2 - 2x - 3 = 0 \Rightarrow x_1 = 3,\ x_2 = -1.$$

 

 

7. $\dfrac{x^2 + 2x + 2}{x + 1} + \dfrac{x^2 + 8x + 20}{x + 4} = \dfrac{x^2 + 4x + 6}{x+2} + \dfrac{x^2 + 6x + 12}{x+3}$

解答:

可参考例题10. $$\frac{(x+1)^2 + 1}{x+1} + \frac{(x+4)^2 + 4}{x+4} = \frac{(x+2)^2 + 2}{x+2} + \frac{(x+3)^2 + 3}{x+3}$$ $$\Rightarrow x+1 + \frac{1}{x+1} + x+4 + \frac{4}{x+4} = x+2 + \frac{2}{x+2} + x+3 + \frac{3}{x+3}$$ $$\Rightarrow \frac{1}{x+1} + \frac{4}{x+4} = \frac{2}{x+2} + \frac{3}{x+3}$$ $$\Rightarrow x_1 = 0,\ x_2 = -\frac{5}{2}.$$ 经检验 $x_1 = 0,\ x_2 = -\dfrac{5}{2}$ 均为方程的解.

 

 

 

主讲教师:

赵胤, 理学硕士(数学) & 教育硕士(数学), 中国数学奥林匹克一级教练员, 高级中学数学教师资格.