Question

\(\dfrac{3x}{2x+4}\)

\(\dfrac{x+3}{x^2-4}\)

\(\dfrac{x+5}{x^2+4x+4}\)

\(\dfrac{x}{3x+6}\)

Answer 1

MTC=6(x+2)^2*(x-2)

\(\dfrac{3x}{2x+4}=\dfrac{3x}{2\left(x+2\right)}=\dfrac{3x\cdot3\cdot\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)^2\cdot\left(x-2\right)}=\dfrac{9x\left(x^2-4\right)}{6\left(x+2\right)^2\left(x-2\right)}\)

\(\dfrac{x+3}{x^2-4}=\dfrac{6\left(x+3\right)\left(x+2\right)}{6\left(x+2\right)^2\cdot\left(x-2\right)}\)

\(\dfrac{x+5}{x^2+4x+4}=\dfrac{\left(x+5\right)}{\left(x+2\right)^2}=\dfrac{6\left(x-2\right)\left(x+5\right)}{6\left(x+2\right)^2\cdot\left(x-2\right)}\)

\(\dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{2x\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)^2\left(x-2\right)}\)