Answer
$\dfrac{\dfrac{4}{x}+\dfrac{x}{x+1}}{\dfrac{1}{2x}+\dfrac{1}{x+6}}=\dfrac{2(x+6)(x+2)}{3(x+1)}$
Work Step by Step
$\dfrac{\dfrac{4}{x}+\dfrac{x}{x+1}}{\dfrac{1}{2x}+\dfrac{1}{x+6}}$
Evaluate the sums indicated in the numerator and the denominator:
$\dfrac{\dfrac{4}{x}+\dfrac{x}{x+1}}{\dfrac{1}{2x}+\dfrac{1}{x+6}}=\dfrac{\dfrac{4(x+1)+x^{2}}{x(x+1
)}}{\dfrac{x+6+2x}{2x(x+6)}}=\dfrac{\dfrac{x^{2}+4x+4}{x(x+1)}}{\dfrac{3x+6}{2x(x+6)}}=...$
Evaluate the division:
$...=\dfrac{x^{2}+4x+4}{x(x+1)}\div\dfrac{3x+6}{2x(x+6)}=\dfrac{2x(x+6)(x^{2}+4x+4)}{x(x+1)(3x+6)}=...$
Factor the trinomial in the numerator and take out common factor $3$ from the second parentheses in the denominator. Then, simplify:
$...=\dfrac{2x(x+6)(x+2)^{2}}{3x(x+1)(x+2)}=\dfrac{2(x+6)(x+2)}{3(x+1)}$
