Question

Tính:

\(a,\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\)

\(b,\dfrac{3-2x}{x^2-9}+\dfrac{1}{2x-6}\)

Answer 1

Lời giải:

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}=\frac{6}{x(x+4)}+\frac{3}{2(x+4)}=\frac{12}{2x(x+4)}+\frac{3x}{2x(x+4)}\)

\(=\frac{12+3x}{2x(x+4)}=\frac{3(x+4)}{2x(x+4)}=\frac{3}{2x}\)

b)

\(\frac{3-2x}{x^2-9}+\frac{1}{2x-6}=\frac{3-2x}{(x-3)(x+3)}+\frac{1}{2(x-3)}\)

\(=\frac{6-4x}{2(x-3)(x+3)}+\frac{x+3}{2(x-3)(x+3)}=\frac{6-4x+x+3}{2(x-3)(x+3)}=\frac{9-3x}{2(x-3)(x+3)}\)

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