a: \(A=\left(x^2-4\right)\cdot\dfrac{3}{x-2}=3\left(x+2\right)\)
b: \(B=\dfrac{8y^2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-\left(x-3\right)}{y}=\dfrac{-8y}{x+3}\)
`A = (x-2)(x+2)sqrt(3/(x-2)^2)`
`= (x+2). sqrt 3`
`B = sqrt((64y^4(x-3)^2)/(-(x-3)(x+3)y^2)`
`= sqrt(-64y^2(x-3))/(x+3))`
`= 8y sqrt ((3-x)/(x+3))`
Với x > 2 ⇔ x - 2 > 0
\(A=\left(x-2\right)\left(x+2\right)\sqrt{\dfrac{9}{\left(x-2\right)^2}}=\left(x-2\right)\left(x+2\right).\dfrac{3}{\left|x-2\right|}=\left(x-2\right)\left(x+2\right)\dfrac{3}{x-2}=3\left(x+2\right)=3x+6\)
\(B=\dfrac{8y^2}{\left(x-3\right)\left(x+3\right)}.\left|\dfrac{x-3}{y}\right|=\dfrac{8y^2}{\left(x-3\right)\left(x+3\right)}.\dfrac{-\left(x-3\right)}{y}\)(vì x - 3 < 0 ; y > 0)
\(=\dfrac{-8y}{x+3}\)
