2016猿辅导初中数学竞赛训练营作业题解答-9

Posted 2020-08-20 赵胤数学课

 

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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 $${1\\over b^2 + c^2 - a^2} + {1\\over c^2 + a^2 - b^2} + {1\\over a^2 + b^2 - c^2}$$ $$= {1\\over -2ab} + {1\\over -2bc} + {1\\over -2ca}= {a + b + c \\over -2abc} = 0.$$