a) ĐKXĐ: \(x\ne0\)
\(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{2\left(4x+1\right)+2x-3}{6x}\)
\(=\dfrac{10x-1}{6x}\)
b) ĐKXĐ: \(x,y\ne0\)
\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right).\left(x+y\right)}{6x^2y^2}.\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)
a) Ta có: \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{2\left(4x+1\right)}{6x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{8x+2+2x-3}{6x}\)
\(=\dfrac{10x-1}{6x}\)
b) Ta có: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{6x^2y^2}\cdot\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)
