Question

 

    mn giúp em ak

 

Answer 1

a: \(=\dfrac{-20xy^2}{45xy}\cdot\dfrac{\left(x-3\right)^2}{x-3}=\dfrac{-4}{5}y\left(x-3\right)\)

b: \(=\dfrac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{5x+15}\)

c: \(=\dfrac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}=\dfrac{x+y-1}{x+1-y}\)

e: \(=\dfrac{\left(x+4+x\right)\left(x+4-x\right)}{2x+4}=4\)

Answer 2

b,\(=\dfrac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\dfrac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{5\left(x+3\right)}\)

Answer 3

d,=\(\dfrac{2x^2+2xy-5x-5y}{2xy-x+2y^2-y}=\dfrac{2x\left(x+y\right)-5\left(x+y\right)}{2y\left(x+y\right)-\left(x+y\right)}=\dfrac{2x-5}{2y-1}\)