$\dfrac{1}{\dfrac{1}{4x}+\dfrac{3}{8y}+\dfrac{3}{4(x+y)}}=\dfrac{1}{\dfrac{2y(x+y)+3x(x+y)+6xy}{8xy(x+y)}}$
$=\dfrac{1}{\dfrac{3x^2+11xy+2y}{8xy(x+y)}}=\dfrac{8xy(x+y)}{3x^2+11xy+2y}$
Rút gọn biều thức $\dfrac{1}{{\dfrac{1}{{4x}} + \dfrac{3}{{8y}} + \dfrac{3}{{4\left( {x + y} \right)}}}}$
$\dfrac{1}{\dfrac{1}{4x}+\dfrac{3}{8y}+\dfrac{3}{4(x+y)}}=\dfrac{1}{\dfrac{2y(x+y)+3x(x+y)+6xy}{8xy(x+y)}}$
$=\dfrac{1}{\dfrac{3x^2+11xy+2y}{8xy(x+y)}}=\dfrac{8xy(x+y)}{3x^2+11xy+2y}$