2016猿辅导初中数学竞赛训练营作业题解答-10

 

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. 化简: $${1\over a-x} - {1\over a+x} - {2x \over a^2 + x^2} - {4x^3 \over a^4 + x^4} - {8x^7 \over a^8 - x^8}$$ 解答: $${1\over a-x} - {1\over a+x} - {2x \over a^2 + x^2} - {4x^3 \over a^4 + x^4} - {8x^7 \over a^8 - x^8}$$ $$= {2x \over a^2 - x^2} - {2x \over a^2 + x^2} - {4x^3 \over a^4 + x^4} - {8x^7 \over a^8 - x^8}$$ $$={4x^3 \over a^4 - x^4} - {4x^3 \over a^4 + x^4} - {8x^7 \over a^8 - x^8} = {8x^7 \over a^8 - x^8} - {8x^7 \over a^8 - x^8} = 0.$$

 

2. 化简: $${x - c \over (x-a)(x-b)} + {b-c \over (a-b)(x-b)} + {b-c \over (b-a)(x-a)}$$ 解答: $${x - c \over (x-a)(x-b)} + {b-c \over (a-b)(x-b)} + {b-c \over (b-a)(x-a)}$$ $$= {x-c \over (x - a)(x - b)} + {(b - c)(x - a) + (b - c)(b - x) \over (x - a)(x - b)(a - b)}$$ $$= {x - c \over (x - a)(x - b)} + {(b - c)(b - a) \over (x - a)(x - b)(a - b)}$$ $$= {x - c \over (x - a)(x - b)} - {b- c \over (x - a)(x - b)} = {1\over x-a}.$$

 

3. 化简: $${a \over (a-b)(a-c)} + {b\over (b-c)(b-a)} + {c\over (c-a)(c-b)}$$ 解答: $${a \over (a-b)(a-c)} + {b\over (b-c)(b-a)} + {c\over (c-a)(c-b)}$$ $$= {a(b - c) + b(c - a) \over (a - b)(a - c)(b - c)} + {c \over (c - a)(c - b)}$$ $$= {c(b - a) \over (a - b)(a - c)(b - c)} + {c \over (c - a)(c - b)}$$ $$= {c \over (c - b)(a - c)} + {c \over (c - a)(c - b)} = 0.$$

 

4. 化简: $${1\over (a-2)(a-1)} + {1\over (a-1)a} + {1\over a(a+1)} + {1\over (a+1)(a+2)}$$ 解答: $${1\over (a-2)(a-1)} + {1\over (a-1)a} + {1\over a(a+1)} + {1\over (a+1)(a+2)}$$ $$= {1\over a - 2} - {1\over a - 1} + {1\over a - 1} - {1 \over a} + {1\over a} - {1 \over a+1} + {1\over a+1} - {1\over a+2}$$ $$= {1\over a-2} - {1\over a+2} = {4 \over a^2 - 4}.$$

 

5. 设 $a$, $b$, $c$ 两两不等, 化简: $${2a - b - c \over a^2 - ab - ac + bc} + {2b - c - a \over b^2 - ab - bc + ac} + {2c - a - b \over c^2 - ac - bc + ab}$$ 解答: $${2a - b - c \over a^2 - ab - ac + bc} + {2b - c - a \over b^2 - ab - bc + ac} + {2c - a - b \over c^2 - ac - bc + ab}$$ $$= {(a - b) + (a - c) \over (a - b)(a - c)} + {(b - a) + (b - c) \over (b - a)(b - c)} + {(c - a)+(c - b) \over (c -a)(c - b)}$$ $$= {1\over a-b} + {1 \over a-c} + {1 \over b - a} + {1\over b - c} + {1 \over c -a} + {1 \over c - b} = 0.$$

 

6. 设 $x$, $y$, $z$ 都是正数, 求证: $${2\over x+y} + {2\over y + z} + {2\over z+x} \ge {9 \over x+y +z}$$ 解答:

采用分析法证明之. $${2\over x+y} + {2\over y + z} + {2\over z+x} \ge {9 \over x+y +z}$$ $$\Leftrightarrow {x + y + z \over x + y} + {x + y + z \over y + z} + {x + y + z \over z + x} \ge {9 \over 2}$$ $$\Leftrightarrow {z \over x + y} + {x \over y + z} + {y \over z + x} \ge {3 \over 2}$$ $$\Leftrightarrow {2z \over x + y} + {2x \over y + z} + {2y \over z + x} \ge 3$$ $$\Leftrightarrow {(x + z)+(y + z) \over  x + y} + {(y + x) + (z + x) \over y+z} + {(z + y)+(x + y) \over z + x} \ge 6$$ $$\Leftrightarrow \left({x + z \over x + y} + {x + y \over z + x}\right) + \left({y + z \over x + y} + {y + x \over y + z}\right) + \left({z + x \over y + z} + {z + y \over z + x}\right) \ge 6$$ 由AM-GM不等式, 最后一个不等式显然成立.

 

7. 分解部分分式: $${12-x \over x(x-3)(x -4)}$$ 解答: $${12-x \over x(x-3)(x -4)} = {A\over x} + {B \over x-3} + {C \over x-4}$$ $$\Rightarrow f(x) = 12 - x = A(x - 3)(x - 4) + Bx(x-4) + Cx(x-3)$$ $$\Rightarrow \begin{cases}f(0) = 12A\\ f(3) = 9 = -3B\\ f(4) = 8 = 4C\end{cases} \Rightarrow \begin{cases}A = 1\\ B = -3\\ C = 2\end{cases}$$ $$\Rightarrow {12-x \over x(x-3)(x -4)} = {1 \over x} - {3 \over x-3} + {2 \over x - 4}.$$