2016猿辅导初中数学竞赛训练营作业题解答-9

 

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1. 若 $\displaystyle{1\over n} - {1\over m} - {1\over n+m} = 0$, 则 $\displaystyle\left({m\over n} + {n \over m}\right)^2 = ?$

解答: $${1\over n} - {1\over m} = {1\over n + m} \Rightarrow m^2 - n^2 = mn$$ $$\Rightarrow {m \over n} - {n \over m} = 1\Rightarrow \left({m \over n} - {n \over m}\right)^2 = 1$$ $$\Rightarrow \left({m \over n} + {n \over m}\right)^2 = \left({m \over n} - {n \over m}\right)^2 + 4 = 5.$$

 

2. 若 $\displaystyle{1\over x} = {2\over y+z} = {3\over z+x}$, 则 $\displaystyle{z+y \over x} = ?$

解答: $${1\over x} = {2\over y+z} \Rightarrow {z+y \over x} = 2.$$

 

3. 若 $\displaystyle{x\over a-b} = {y \over b-c} = {z \over c-a}$, 则 $x + y + z = ?$

解答:

令 $\displaystyle{x\over a-b} = {y \over b-c} = {z \over c-a} = k$, $$\Rightarrow x + y + z = k(a - b + b - c + c - a) = 0.$$

 

4. 若 $x^2 - 8x + 13 = 0$, 则 $\displaystyle{x^4 - 6x^3 - 2x^2 + 18x + 23 \over x^2 - 8x + 15} = ?$

解答: $$x^4 - 6x^3 - 2x^2 + 18x + 23$$ $$= x^2(x^2 - 8x + 13) + 2x^3 - 15x^2 + 18x + 23$$ $$= 2x(x^2 - 8x + 13) + x^2 - 8x + 23$$ $$= x^2 - 8x + 13 + 10 = 10,$$ $$x^2 - 8x + 15 = x^2 - 8x + 13 + 2 = 2,$$ $$\Rightarrow {x^4 - 6x^3 - 2x^2 + 18x + 23 \over x^2 - 8x + 15} = 5.$$

 

5. 若 $x + y + z = 3$, 且 $x, y, z$ 不全相等, 则 $\displaystyle{3(x-1)(y-1)(z-1) \over (x-1)^3 + (y-1)^3 + (z-1)^3} = ?$

解答: $$x+y+z = 3 \Rightarrow (x - 1)+(y - 1)+(z - 1) = 0$$ $$\Rightarrow (x - 1)^3 + (y - 1)^3 + (z - 1)^3 = 3(x - 1)(y - 1)(z - 1)$$ $$\Rightarrow {3(x-1)(y-1)(z-1) \over (x-1)^3 + (y-1)^3 + (z-1)^3} = 1.$$

 

6. 若 $\displaystyle{x \over y} = {5\over3}$, $\displaystyle{y\over z} = {7 \over 2}$, 则 $\displaystyle{x-y \over y-z} =?$

解答: $$x = {5\over3}y,\ z = {2\over7}y,$$  $$\Rightarrow {x-y \over y-z} = {{2\over3}y\over{5\over7}y} = {14\over15}.$$

 

7. 若 $\displaystyle{3\over x} - {2\over y} = 5$, 则 $\displaystyle{2x + 7xy - 3y \over 4x - xy - 6y} =?$

解答: $${2x + 7xy - 3y \over 4x - xy - 6y}$$ $$= {{2\over y} + 7 - {3\over x} \over {4\over y} - 1 - {6 \over x}}$$ $$= {7 - 5 \over -1 - 10} = -{2\over 11}.$$

 

8. 若 $\displaystyle x + {1\over x} = a$, 则 $\displaystyle x^6 + {1\over x^6} = ?$

解答: $$x^2 + {1\over x^2} = \left(x + {1\over x}\right)^2 - 2 = a^2 - 2,$$ $$x^4 + {1\over x^4} = \left(x^2 + {1\over x^2}\right)^2 - 2 = a^4 - 4a^2 + 2,$$ $$\Rightarrow x^6 + {1\over x^6} = \left(x^2 + {1\over x^2}\right)\left(x^4 + {1\over x^4} - 1\right)$$ $$= (a^2 - 2)(a^4 - 4a^2 + 1) = a^6 - 6a^4 + 9a^2 - 2.$$

 

9. 若 $\displaystyle x+ {1\over x} = -4$, 则 $\displaystyle x^3 + {1\over x^3} = ?$

解答: $$x^3 + {1\over x^3} = \left(x + {1\over x}\right)\left(x^2 + {1\over x^2} - 1\right)$$ $$= \left(x + {1\over x}\right)\left[\left(x + {1\over x}\right)^2 - 3\right] = -52.$$

 

10. 若 $(x-2y + 2)^2 = x(y-1)$, 则 $\displaystyle{x\over y-1} = ?$

解答: $$(x-2y + 2)^2 = x(y-1)$$ $$\Rightarrow x^2 + 4(y-1)^2 - 4x(y-1) = x(y-1)$$ $$\Rightarrow x^2 - 5x(y-1) + 4(y - 1)^2 = 0$$ $$\Rightarrow \left[x - (y - 1)\right]\left[x - 4(y-1)\right] = 0$$ $$\Rightarrow x_1 = y-1,\ x_2 = 4(y - 1).$$ 因此原式的值为 $1$ 或 $4$.

 

11. 若 $x^2 - 3x + 1 = 0$, 则 $\displaystyle{2x^5 - 5x^4 + 2x^3 - 8x^2 \over x^2 + 1} = ?$

解答: $$2x^5 - 5x^4 + 2x^3 - 8x^2 = 2x^3(x^2 - 3x + 1) + x^4 - 8x^2$$ $$= x^2(x^2 - 3x + 1) + 3x^3 - 9x^2$$ $$= 3x(x^2 - 3x + 1) - 3x$$ $$= -3x = -(x^2 + 1)$$ $$\Rightarrow {2x^5 - 5x^4 + 2x^3 - 8x^2 \over x^2 + 1} = -1.$$

 

12. 若 $a, b, c$ 均不为0, 且 $a+b+c = 0$, 则 $\displaystyle{1\over b^2 + c^2 - a^2} + {1\over c^2 + a^2 - b^2} + {1\over a^2 + b^2 - c^2} = ?$

解答: $$a + b + c = 0\Rightarrow a + b = -c$$ $$\Rightarrow a^2 + b^2 - c^2 = a^2 + b^2 - (a + b)^2 = -2ab.$$ 同理可得 $$b^2 + c^2 - a^2 = -2bc,\ c^2 + a^2 - b^2 = -2ca.$$ 因此 $${1\over b^2 + c^2 - a^2} + {1\over c^2 + a^2 - b^2} + {1\over a^2 + b^2 - c^2}$$ $$= {1\over -2ab} + {1\over -2bc} + {1\over -2ca}= {a + b + c \over -2abc} = 0.$$