\(a,\dfrac{3x}{2x+4}=\dfrac{3x}{2\left(x+2\right)}=\dfrac{3x\left(x-2\right)}{2\left(x^2-4\right)}=\dfrac{3x^2-6x}{2\left(x^2-4\right)}\)
\(\dfrac{x+3}{x^2-4}=\dfrac{2\left(x+3\right)}{2\left(x^2-4\right)}=\dfrac{2x+6}{2\left(x^2-4\right)}\)
\(b,\dfrac{x+5}{x^2+4x+4}=\dfrac{x+5}{\left(x+2\right)^2}=\dfrac{3\left(x+5\right)}{3\left(x+2\right)^2}=\dfrac{3x+15}{3\left(x+2\right)^2}\\ \dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}=\dfrac{x^2+2x}{3\left(x+2\right)^2}\)
Lời giải:
a.
\(\frac{3x}{2x+4}=\frac{3x(x-2)}{2(x+2)(x-2)}=\frac{3x^2-6x}{2(x^2-4)}\)
\(\frac{x+3}{x^2-4}=\frac{2(x+3)}{2(x^2-4)}\)
b.
\(\frac{x+5}{x^2+4x+4}=\frac{x+5}{(x+2)^2}=\frac{3(x+5)}{3(x+2)^2}\)
\(\frac{x}{3x+6}=\frac{x}{3(x+2)}=\frac{x(x+2)}{3(x+2)^2}\)
