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: $${1over a-x} - {1over a+x} - {2x over a^2 + x^2} - {4x^3 over a^4 + x^4} - {8x^7 over a^8 - x^8}$$ 解答: $${1over a-x} - {1over a+x} - {2x over a^2 + x^2} - {4x^3 over a^4 + x^4} - {8x^7 over a^8 - x^8}$$ $$= {2x over a^2 - x^2} - {2x over a^2 + x^2} - {4x^3 over a^4 + x^4} - {8x^7 over a^8 - x^8}$$ $$={4x^3 over a^4 - x^4} - {4x^3 over a^4 + x^4} - {8x^7 over a^8 - x^8} = {8x^7 over a^8 - x^8} - {8x^7 over a^8 - x^8} = 0.$$
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: $${x - c over (x-a)(x-b)} + {b-c over (a-b)(x-b)} + {b-c over (b-a)(x-a)}$$ $$= {x-c over (x - a)(x - b)} + {(b - c)(x - a) + (b - c)(b - x) over (x - a)(x - b)(a - b)}$$ $$= {x - c over (x - a)(x - b)} + {(b - c)(b - a) over (x - a)(x - b)(a - b)}$$ $$= {x - c over (x - a)(x - b)} - {b- c over (x - a)(x - b)} = {1over x-a}.$$
3. 化简: $${a over (a-b)(a-c)} + {bover (b-c)(b-a)} + {cover (c-a)(c-b)}$$ 解答: $${a over (a-b)(a-c)} + {bover (b-c)(b-a)} + {cover (c-a)(c-b)}$$ $$= {a(b - c) + b(c - a) over (a - b)(a - c)(b - c)} + {c over (c - a)(c - b)}$$ $$= {c(b - a) over (a - b)(a - c)(b - c)} + {c over (c - a)(c - b)}$$ $$= {c over (c - b)(a - c)} + {c over (c - a)(c - b)} = 0.$$
4. 化简: $${1over (a-2)(a-1)} + {1over (a-1)a} + {1over a(a+1)} + {1over (a+1)(a+2)}$$ 解答: $${1over (a-2)(a-1)} + {1over (a-1)a} + {1over a(a+1)} + {1over (a+1)(a+2)}$$ $$= {1over a - 2} - {1over a - 1} + {1over a - 1} - {1 over a} + {1over a} - {1 over a+1} + {1over a+1} - {1over a+2}$$ $$= {1over a-2} - {1over a+2} = {4 over a^2 - 4}.$$
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: $${2a - b - c over a^2 - ab - ac + bc} + {2b - c - a over b^2 - ab - bc + ac} + {2c - a - b over c^2 - ac - bc + ab}$$ $$= {(a - b) + (a - c) over (a - b)(a - c)} + {(b - a) + (b - c) over (b - a)(b - c)} + {(c - a)+(c - b) over (c -a)(c - b)}$$ $$= {1over a-b} + {1 over a-c} + {1 over b - a} + {1over b - c} + {1 over c -a} + {1 over c - b} = 0.$$
6. 设 $x$, $y$, $z$ 都是正数, 求证: $${2over x+y} + {2over y + z} + {2over z+x} ge {9 over x+y +z}$$ 解答:
采用分析法证明之. $${2over x+y} + {2over y + z} + {2over z+x} ge {9 over x+y +z}$$ $$Leftrightarrow {x + y + z over x + y} + {x + y + z over y + z} + {x + y + z over z + x} ge {9 over 2}$$ $$Leftrightarrow {z over x + y} + {x over y + z} + {y over z + x} ge {3 over 2}$$ $$Leftrightarrow {2z over x + y} + {2x over y + z} + {2y over z + x} ge 3$$ $$Leftrightarrow {(x + z)+(y + z) over x + y} + {(y + x) + (z + x) over y+z} + {(z + y)+(x + y) over z + x} ge 6$$ $$Leftrightarrow left({x + z over x + y} + {x + y over z + x} ight) + left({y + z over x + y} + {y + x over y + z} ight) + left({z + x over y + z} + {z + y over z + x} ight) ge 6$$ 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: $${12-x over x(x-3)(x -4)} = {Aover x} + {B over x-3} + {C over x-4}$$ $$Rightarrow f(x) = 12 - x = A(x - 3)(x - 4) + Bx(x-4) + Cx(x-3)$$ $$Rightarrow egin{cases}f(0) = 12A\ f(3) = 9 = -3B\ f(4) = 8 = 4Cend{cases} Rightarrow egin{cases}A = 1\ B = -3\ C = 2end{cases}$$ $$Rightarrow {12-x over x(x-3)(x -4)} = {1 over x} - {3 over x-3} + {2 over x - 4}.$$