Question

奥数计算题

1.(x^3-x^2-4x+1)/(x^2-3x+2)-(x^3-2x^2-9x+21)/（x^2-5x+6）+（x^2-3x+8）/（x^2-4x+3）
2.{(x^2)/（x-y)}*{y/(x+y)}-{(x^4y)/(x^4-y^4)}÷{（x^2/x^2+y^2）^2}

Answer 1

1.将3个分数化成6个，当然其中要有一定的技巧分配。
看分母是(x-1)(x-2)(x-3)的组合，所以想到将分子也配成这样的来化简：[(x^3-4x)-(x^2-1)]/[(x-2)(x-1)] - [(x^3-4x^2+4x)+(2x^2-13x+21)]/[(x-2)(x-3)] + [(x^2-7x+12)+(4x-4)]/[x-3)(x-1)]
= [x(x+2)(x-2)-(x+1)(x-1)]/[(x-2)(x-1)] - [x(x-2)^2+(2x-7)(x-3)]/[(x-2)(x-3)] + [(x-3)(x-4)+4(x-1)]/[x-3)(x-1)]
想式子化简成6个只有1个分母的式子：[x(x+2)]/(x-1) - (x+1)/(x-2) -[x(x-2)]/(x-3) - (2x-7)/(x-2) + (x-4)/(x-1) + 4/(x-3)
再将分母相同的组合在一起：[x(x+2)+(x-4)]/(x-1) - [(x+1)+(2x-7)]/(x-2) -[x(x-2)-4]/(x-3)
=(x^2+3x-4)/(x-1) - (3x-6)/(x-2) - (x^2-2x-4)/(x-3)
=[(x-1)(x+4)]/(x-1) - [3(x-2)]/(x-2) - [(x^2-2x-3)-1]/(x-3)
=(x+4)- 3 - [(x+1)(x-3)-1]/(x-3)
=x+1 - {[(x+1)(x-3)]/(x-3)-1/(x-3)}
=x+1-{(x+1)-1/(x-3)
=1/(x-3)
2. 这个题觉得你写的可能会出现歧义，我是以以下式子解答：
{(x^2)/(x-y)}*{y/(x+y)}-{(y*x^4)/(x^4-y^4)}÷{(x^2)/(x^2+y^2)^2}
=(y*x^2)/(x^2-y^2) - {(y*x^4)/[(x^2+y^2)(x^2-y^2)]} × {[(x^2+y^2)^2]/(x^4)}
对后面式子进行化简，主要是约分(如果这样不好看 请自己将上式用分数写下来)：
=(y*x^2)/(x^2-y^2) - {y(x^2+y^2)}/(x^2-y^2)
=(y*x^2)/(x^2-y^2) - {(y*x^2)/(x^2-y^2)+(y^3)/(x^2-y^2)}
= -(y^3)/(x^2-y^2)