a: \(=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{10x-15}{2x+3}\)b: \(=\dfrac{8xy}{12x^3}\cdot\dfrac{\left(3x-1\right)^3}{-\left(3x-1\right)}=\dfrac{-2y\left(3x-1\right)^2}{3x^2}\)c: \(=\dfrac{15x^2y}{35x^3y^2}\cdot\dfrac{\left(x-2y\right)^2}{\left(x-2y\right)^3}\cdot\left(-1\right)=\dfrac{-3}{7xy\cdot\left(x-2y\right)}\)d: \(=\dfrac{10xy^2}{12x^3}\cdot\dfrac{\left(2x-1\right)^3}{2x-1}=\dfrac{5y^2\left(2x-1\right)^2}{6x^2}\)e: \(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}=\dfrac{x+y}{x+y+z}\)f: \(=\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3\left(x+3\right)}{x-1}\)Câu trả lời: a: \(=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{10x-15}{2x+3}\)b: \(=\dfrac{8xy}{12x^3}\cdot\dfrac{\left(3x-1\right)^3}{-\left(3x-1\right)}=\dfrac{-2y\left(3x-1\right)^2}{3x^2}\)c: \(=\dfrac{15x^2y}{35x^3y^2}\cdot\dfrac{\left(x-2y\right)^2}{\left(x-2y\right)^3}\cdot\left(-1\right)=\dfrac{-3}{7xy\cdot\left(x-2y\right)}\)d: \(=\dfrac{10xy^2}{12x^3}\cdot\dfrac{\left(2x-1\right)^3}{2x-1}=\dfrac{5y^2\left(2x-1\right)^2}{6x^2}\)e: \(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}=\dfrac{x+y}{x+y+z}\)f: \(=\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3\left(x+3\right)}{x-1}\)