Question

分式的加减法(x+2y)+(4y^2)/(x-2y)-4x^2y/x^2-4y^2x^3+1/(x^3-2x^2+2x-

Answer 1

(x+2y)+(4y^2)/(x-2y)-4x^2y/x^2-4y^2

=[(x+2y)^2(x-2y)+4y^2(x+2y)-4x^2y]/(x+2y)(x-2y)

=[(x+2y)(x^2-4y^2)+4xy^2+8y^3-4x^2y]/(x^2-4y^2)

=(x^3-4xy^2+2x^2y-8y^3+4xy^2+8y^3-4x^2y)/(x^2-4y^2)

=(x^3-2x^2y)/(x^2-4y^2)

=x^2(x-2y)/(x+2y)(x-2y)

=x^2/(x+2y)

x^3+1/(x^3-2x^2+2x-1)+(x^3-1)/(x^3+2x^2+2x+1)-(x^2+1)/(x^2-1)

=(x^3+1)/[x(x^2-2x+1)+(x-1)]+(x^3-1)/[x(x^2+2x+1)+(x+1)]-(x^2+1)/(x^2-1)

=(x+1)(x^2-x+1)/[x(x-1)^2+(x-1)]+(x^3-1)/[x(x+1)^2+(x+1)]-(x^2+1)/(x^2-1)

=(x+1)(x^2-x+1)/[(x-1)(x^2-x+1)]+(x-1)(x^2+x+1)/[(x+1)(x^2+x+1)-(x^2+1)/(x^2-1)

=(x+1)/(x-1)+(x-1)/(x+1)-(x^2+1)/(x^2-1)

=[(x+1)^2+(x-1)^2-(x^2+1)]/(x^2-1)

=(x^2+2x+1+x^2-2x+1-x^2-1)/(x^2-1)

=(x^2+1)/(x^2-1)

做得好累啊!