扫描以下二维码下载并安装猿辅导App, 打开后请搜索教师姓名"赵胤"即可报名本课程(14次课, 99元).
1. 若 $displaystyle{1over n} - {1over m} - {1over n+m} = 0$, 则 $displaystyleleft({mover n} + {n over m} ight)^2 = ?$
解答: $${1over n} - {1over m} = {1over n + m} Rightarrow m^2 - n^2 = mn$$ $$Rightarrow {m over n} - {n over m} = 1Rightarrow left({m over n} - {n over m} ight)^2 = 1$$ $$Rightarrow left({m over n} + {n over m} ight)^2 = left({m over n} - {n over m} ight)^2 + 4 = 5.$$
2. 若 $displaystyle{1over x} = {2over y+z} = {3over z+x}$, 则 $displaystyle{z+y over x} = ?$
解答: $${1over x} = {2over y+z} Rightarrow {z+y over x} = 2.$$
3. 若 $displaystyle{xover a-b} = {y over b-c} = {z over c-a}$, 则 $x + y + z = ?$
解答:
令 $displaystyle{xover a-b} = {y over b-c} = {z over c-a} = k$, $$Rightarrow x + y + z = k(a - b + b - c + c - a) = 0.$$
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 $displaystyle{x over y} = {5over3}$, $displaystyle{yover z} = {7 over 2}$, 则 $displaystyle{x-y over y-z} =?$
解答: $$x = {5over3}y, z = {2over7}y,$$ $$Rightarrow {x-y over y-z} = {{2over3}yover{5over7}y} = {14over15}.$$
7. 若 $displaystyle{3over x} - {2over y} = 5$, 则 $displaystyle{2x + 7xy - 3y over 4x - xy - 6y} =?$
解答: $${2x + 7xy - 3y over 4x - xy - 6y}$$ $$= {{2over y} + 7 - {3over x} over {4over y} - 1 - {6 over x}}$$ $$= {7 - 5 over -1 - 10} = -{2over 11}.$$
8. 若 $displaystyle x + {1over x} = a$, 则 $displaystyle x^6 + {1over x^6} = ?$
解答: $$x^2 + {1over x^2} = left(x + {1over x} ight)^2 - 2 = a^2 - 2,$$ $$x^4 + {1over x^4} = left(x^2 + {1over x^2} ight)^2 - 2 = a^4 - 4a^2 + 2,$$ $$Rightarrow x^6 + {1over x^6} = left(x^2 + {1over x^2} ight)left(x^4 + {1over x^4} - 1 ight)$$ $$= (a^2 - 2)(a^4 - 4a^2 + 1) = a^6 - 6a^4 + 9a^2 - 2.$$
9. 若 $displaystyle x+ {1over x} = -4$, 则 $displaystyle x^3 + {1over x^3} = ?$
解答: $$x^3 + {1over x^3} = left(x + {1over x} ight)left(x^2 + {1over x^2} - 1 ight)$$ $$= left(x + {1over x} ight)left[left(x + {1over x} ight)^2 - 3 ight] = -52.$$
10. 若 $(x-2y + 2)^2 = x(y-1)$, 则 $displaystyle{xover y-1} = ?$
解答: $$(x-2y + 2)^2 = x(y-1)$$ $$Rightarrow x^2 + 4(y-1)^2 - 4x(y-1) = x(y-1)$$ $$Rightarrow x^2 - 5x(y-1) + 4(y - 1)^2 = 0$$ $$Rightarrow left[x - (y - 1) ight]left[x - 4(y-1) ight] = 0$$ $$Rightarrow x_1 = y-1, x_2 = 4(y - 1).$$ 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 $displaystyle{1over b^2 + c^2 - a^2} + {1over c^2 + a^2 - b^2} + {1over a^2 + b^2 - c^2} = ?$
解答: $$a + b + c = 0Rightarrow a + b = -c$$ $$Rightarrow a^2 + b^2 - c^2 = a^2 + b^2 - (a + b)^2 = -2ab.$$ 同理可得 $$b^2 + c^2 - a^2 = -2bc, c^2 + a^2 - b^2 = -2ca.$$ 因此 $${1over b^2 + c^2 - a^2} + {1over c^2 + a^2 - b^2} + {1over a^2 + b^2 - c^2}$$ $$= {1over -2ab} + {1over -2bc} + {1over -2ca}= {a + b + c over -2abc} = 0.$$