\(M=\left(-2x^2y\right)>.\left(-\dfrac{1}{2}xy^2\right)^2\)
\(M=\left(-2x^2y\right).\left(-\dfrac{1}{2}\right)^2.x^2.\left(y^2\right)^2\)
\(M=\left(-2x^2y\right).\dfrac{1}{4}x^2y^4\)
\(M=\left(-2.\dfrac{1}{4}\right).\left(x^2.x^2\right).\left(y.y^4\right)\)
\(M=\dfrac{-1}{2}x^4y^5\)
Ta thay \(x=\dfrac{1}{2};y=-1\) vào đơn thức \(M=\dfrac{-1}{2}x^4y^5\) ta được:
\(M=\dfrac{-1}{2}.\left(-\dfrac{1}{2}\right)^4.\left(-1\right)^5\)
\(M=\dfrac{-1}{2}.\dfrac{1}{16}.\left(-1\right)\)
\(M=\dfrac{1}{32}\)
\(M=\left(-2x^2y\right).\left(\dfrac{-1}{2}xy^2\right)^2\)
\(=\left(-2x^2y\right).\left(\dfrac{1}{4}x^2y^4\right)\)
\(=\left(-2.\dfrac{1}{4}\right).\left(x^2x^2\right).\left(yy^4\right)\)
\(=\dfrac{-1}{2}x^4y^5\)
Thay \(x=\dfrac{1}{2};y=-1\) vào M ta được:
\(M=\dfrac{-1}{2}.\left(\dfrac{1}{2}\right)^4.\left(-1\right)^5\)
\(=\dfrac{-1}{2}.\dfrac{1}{16}.\left(-1\right)\)
\(=\dfrac{1}{32}\)
Vậy M \(=\dfrac{1}{32}\) tại \(x=\dfrac{1}{2};y=-1\)
y
a) ( -2xy2 )3.(-3xy) b) (-3xy2)2. 1 xy c) (-2x).(-0.5xyz)
Bài 7: Rút gọn rồi tính giá trị biểu thức A = 2x2y – 3xy2 – x2y + 2xy2 –xy + 1 tại x = -2; y = 1