Bài toán khó

Bài 4: a) Ta có: \[ A = 2x^2y^2(x^3y^2 - x^2y^3 - \frac{1}{2}y^5) \] \[ = 2x^2y^2 \cdot x^3y^2 - 2x^2y^2 \cdot x^2y^3 - 2x^2y^2 \cdot \frac{1}{2}y^5 \] \[ = 2x^{2+3}y^{2+2} - 2x^{2+2}y^{2+3} - 2x^2y^2 \cdot \frac{1}{2}y^5 \] \[ = 2x^5y^4 - 2x^4y^5 - x^2y^2 \cdot y^5 \] \[ = 2x^5y^4 - 2x^4y^5 - x^2y^7 \] b) Ta có: \[ B = -\frac{1}{3}xy(3x^3y^2 - 6x^2 + y^2) \] \[ = -\frac{1}{3}xy \cdot 3x^3y^2 - \frac{1}{3}xy \cdot (-6x^2) - \frac{1}{3}xy \cdot y^2 \] \[ = -\frac{1}{3}xy \cdot 3x^3y^2 + \frac{1}{3}xy \cdot 6x^2 - \frac{1}{3}xy \cdot y^2 \] \[ = -x^4y^3 + 2x^3y - \frac{1}{3}xy^3 \] c) Ta có: \[ C = (-2xy^2 + \frac{2}{3}y^2 + 4xy^2) \cdot \frac{3}{2}xy \] \[ = (-2xy^2 \cdot \frac{3}{2}xy) + (\frac{2}{3}y^2 \cdot \frac{3}{2}xy) + (4xy^2 \cdot \frac{3}{2}xy) \] \[ = -3x^2y^3 + xy^3 + 6x^2y^3 \] \[ = 3x^2y^3 + xy^3 \] d) Ta có: \[ D = (x^2 + 2xy - 3)(-xy) \] \[ = x^2 \cdot (-xy) + 2xy \cdot (-xy) - 3 \cdot (-xy) \] \[ = -x^3y - 2x^2y^2 + 3xy \] e) Ta có: \[ E = \frac{1}{2}x^2y(2x^3 - \frac{2}{5}xy^2 - 1) \] \[ = \frac{1}{2}x^2y \cdot 2x^3 - \frac{1}{2}x^2y \cdot \frac{2}{5}xy^2 - \frac{1}{2}x^2y \cdot 1 \] \[ = x^5y - \frac{1}{5}x^3y^3 - \frac{1}{2}x^2y \] f) Ta có: \[ F = (-xy^2)^2 \cdot (x^2 - 2x + 1) \] \[ = x^2y^4 \cdot (x^2 - 2x + 1) \] \[ = x^2y^4 \cdot x^2 - x^2y^4 \cdot 2x + x^2y^4 \cdot 1 \] \[ = x^4y^4 - 2x^3y^4 + x^2y^4 \]