Bài 2:
a) \(A=\left(\dfrac{2}{3}x^3y^3\right).\left(\dfrac{1}{2}xy^2\right)^3.\left(\dfrac{-8}{3}x^2\right)\)
\(A=\left(\dfrac{2}{3}x^3y^3\right).\left(\dfrac{1}{2}\right)^3.x^3.\left(y^2\right)^3.\left(\dfrac{-8}{3}x^2\right)\)
\(A=\left(\dfrac{2}{3}x^3y^3\right).\dfrac{1}{8}.x^3.y^6.\left(\dfrac{-8}{3}x^2\right)\)
\(A=\left[\dfrac{2}{3}.\dfrac{1}{8}.\left(\dfrac{-8}{3}\right)\right].\left(x^3.x^3.x^2\right).\left(y^3.y^6\right)\)
\(A=\dfrac{-2}{9}x^8y^9\)
b) Tại \(x=2\) ; \(y=\dfrac{-1}{2}\) ta có:
\(A=\dfrac{-2}{9}.2^8.\left(\dfrac{-1}{2}\right)^9\)
\(A=\dfrac{-2}{9}.256.\left(\dfrac{-1}{512}\right)=\dfrac{1}{9}\)
Vậy tại \(x=2\) ; \(y=\dfrac{-1}{2}\) vào biểu thức A là \(\dfrac{1}{9}\)
a,- (\(\dfrac{2}{3}\)x3y3).(\(\dfrac{1}{2}\)xy2)3.(\(\dfrac{-8}{5}\)x2)
=\(\dfrac{2}{3}\)x3y3.\(\dfrac{1}{2}\)x3y6.\(\dfrac{-8}{5}\)x2
=(\(\dfrac{2}{3}\).\(\dfrac{1}{2}\).\(\dfrac{-8}{5}\)).(x3x3x2).(y3y6)
=\(\dfrac{-8}{15}\)x8y9
- Biến: x8y9
- Bậc: 17
