2016猿辅导初中数学竞赛训练营作业题解答-10

Posted 2020-08-21 赵胤数学课

 

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