Question

Thực hiện phép tính và rút gọn:
a) \(\dfrac{x-2}{6x^2-6x}-\dfrac{1}{4x^2-4}\)
b) \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{6x^3+6}:\dfrac{x^2-1}{4x^2-4x+4}\)

Answer 1

Lời giải:

a)

\(\frac{x-2}{6x^2-6x}-\frac{1}{4x^2-4}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x^2-1)}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x-1)(x+1)}\)

\(=\frac{2(x+1)(x-2)}{12x(x-1)(x+1)}-\frac{3x}{12x(x-1)(x+1)}=\frac{2(x+1)(x-2)-3x}{12x(x-1)(x+1)}\)

\(=\frac{2x^2-5x-4}{12x(x-1)(x+1)}=\frac{2x^2-5x-4}{12x^3-12x}\)

b) ĐK: \(x\neq \pm 1\)

\(\frac{(x+1)(x^2-2x+1)}{6x^3+6}:\frac{x^2-1}{4x^2-4x+4}\)

\(=\frac{(x+1)(x-1)^2}{6(x^3+1)}.\frac{4x^2-4x+4}{x^2-1}\)

\(=\frac{4(x+1)(x-1)^2(x^2-x+1)}{6(x+1)(x^2-x+1)(x^2-1)}\)

\(=\frac{2(x-1)}{3(x+1)}\)