Question

x²-4/9-y² : x-2/3 + y-2/3-y

Answer 1

\(\dfrac{x^2-4}{9-y^2}:\dfrac{x-2}{3}+\dfrac{y-2}{3-y}\)

ĐKXĐ: \(x\ne\pm3\)

\(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(3-y\right)\left(3+y\right)}.\dfrac{3}{x-2}+\dfrac{-2+y}{3-y}\)

\(=\dfrac{3\left(x+2\right)}{\left(3-y\right)\cdot\left(3+y\right)}+\dfrac{1-\left(3-y\right)}{3-y}\)

\(=\dfrac{3\left(x+2\right)}{\left(3-y\right)\left(3+y\right)}+\dfrac{1}{3-y}-1\)

\(=\dfrac{3\left(x+2\right)}{\left(3-y\right)\left(3+y\right)}-\dfrac{3+y}{\left(3-y\right)\left(3+y\right)}-1\)

\(=\dfrac{3x+6-3-y}{\left(3-y\right)\left(3+y\right)}-1\)

\(=\dfrac{3x-y+3}{\left(3-y\right)\left(3+y\right)}-1\)