a: Đặt \(A=\dfrac{1}{2}x^2\left(-2x^2y^3z\right)\cdot\left(-\dfrac{1}{3}x^2y^3\right)\)
\(=\dfrac{1}{2}\cdot\left(-2\right)\cdot\left(-\dfrac{1}{3}\right)\cdot x^2\cdot x^2\cdot x^2\cdot y^3\cdot y^3\cdot z\)
\(=\dfrac{1}{3}x^6y^6z\)
Khi x=1/2 và y=2 thì \(A=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}\right)^6\cdot2^6\cdot z=\dfrac{1}{3}z\)
b: \(B=\left(-x^2y\right)^3\cdot\dfrac{1}{2}x^2y^3\cdot\left(-2xy^2z\right)^3\)
\(=\left(-1\right)\cdot x^6y^3\cdot\dfrac{1}{2}x^2y^3\left(-8\right)x^3y^6z^3\)
\(=4x^{11}y^{12}z^3\)
Khi x=1/2 và y=2 thì \(B=4\cdot\left(\dfrac{1}{2}\right)^{11}\cdot2^{12}\cdot z^3=4\cdot2\cdot z^3=8z^3\)
c: Đặt \(C=\left(-6x^3yz\right)\cdot\left(\dfrac{2}{3}x^2y\right)^2\)
\(=-6\cdot x^3yz\cdot\dfrac{4}{9}x^4y^2\)
\(=-\dfrac{8}{3}x^7y^3z\)
Khi x=1/2 và y=2 thì \(C=-\dfrac{8}{3}\cdot\left(\dfrac{1}{2}\right)^7\cdot2^3\cdot z=-\dfrac{8}{3}\cdot\dfrac{1}{128}\cdot8\cdot z=-\dfrac{1}{6}z\)
