\(P=\frac{1}{3}x^2y+xy^2-xy+\frac{1}{2}xy^2-5xy-\frac{1}{3}x^2y\)
\(P=\left(\frac{1}{3}x^2y-\frac{1}{3}x^2y\right)+\left(xy^2+\frac{1}{2}xy^2\right)+\left(-xy-5xy\right)\)
\(P=0+\frac{3}{2}xy^2+\left(-6xy\right)=\frac{3}{2}xy^2-6xy\)
Thay x=0,5;y=1 vào P ta có:
\(\frac{3}{2}.0,5.1^2-6.0,5.1=\frac{3}{4}-3=\frac{3}{4}-\frac{12}{4}=-\frac{9}{4}=-2,25\)
Vậy P=-9/4=-2,25