Question

tính giá trị biểu thức 3x^3-2y^3-6x^2y^2+xy taị x=2/3, y=1/2

Answer 1

Ta có: \(3x^3-2y^3-6x^2y^2+xy\)

\(=\left(3x^3-6x^2y^2\right)+\left(xy-2y^3\right)\)

\(=3x^2\left(x-2y^2\right)+y\left(x-2y^2\right)\)

\(=\left(x-2y^2\right)\left(3x^2+y\right)\)

\(=\left(\dfrac{2}{3}-2\cdot\dfrac{1}{4}\right)\cdot\left(3\cdot\dfrac{4}{9}+\dfrac{1}{2}\right)\)

\(=\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\cdot\left(\dfrac{4}{3}+\dfrac{1}{2}\right)\)

\(=\dfrac{1}{6}\cdot\dfrac{11}{6}=\dfrac{11}{36}\)