Bài 1:
\(a,A=x^2y^3-\dfrac{5}{4}\cdot\dfrac{4}{5}x^2y^3=x^2y^3-x^2y^3=0\\ b,B=\dfrac{1}{2}a^3b^2-\dfrac{2}{3}a^3b^2=-\dfrac{1}{6}a^3b^2\)
Bài 2:
\(A=-3x^3y^3z-5x^3y^3z=-8x^3y^3z\)
Bài 3:
\(=-3x^2y^2+4x^2y^2=x^2y^2\)
Vì \(x^2\ge0;y^2\ge0\Leftrightarrow x^2y^2\ge0\left(đpcm\right)\)