Đáp án:
a, \(\frac{6.x^3y^2-2x^4y+2x^2y^2}{3}\)
b, \(\frac{4}{x+2}\)
c, \(\frac{4}{x+2}\)
d, \(\frac{4}{3(x+4)}\)
Giải thích các bước giải:
a, \(\frac{6.x^3y^2-2x^4y+2x^2y^2}{3}\)
b, \(2x^3 - 3x^2 + 10x - 15x - 8x + 12 = 2x^3 - 3x^2 - 13x + 12\)
c, \(\frac{2}{x+2}-\frac{8}{x^2-4}+\frac{2}{x-2}=\frac{2(x-2)-8+2(x+2)}{x^2-4} = \frac{4x-8}{x^2-4}= \frac{4(x-2)}{(x-2).(x+2)}= \frac{4}{x+2}\)
d, \(\frac{4(x+3)}{(x+4)^2}.\frac{(x+4)}{3(x+3)}=\frac{4}{3(x+4)}\)
