\(P=\frac{1}{3}x^2y+xy^2-xy+\frac{1}{2}xy^2-5xy-\frac{1}{3}x^2y\)
\(=\left(\frac{1}{3}x^2y-\frac{1}{3}x^2y\right)+\left(xy^2+\frac{1}{2}xy^2\right)-\left(xy+5xy\right)\)
\(=0+xy^2\left(1+\frac{1}{2}\right)-xy\left(1+5\right)\)
\(=\frac{3}{2}xy^2-6xy\)
Thay x = 0,5 và y = 1 vào đa thức trên ta có:
\(\frac{3}{2}.0,5.1^2-6.0,5.1\)
\(=\frac{3}{2}.\frac{1}{2}.1-6.\frac{1}{2}.1\)
\(=\frac{3}{4}-\frac{6}{2}=\frac{3}{4}-\frac{12}{4}=-\frac{9}{4}\)
P/s: Ko chắc!
P=\(\frac{1}{3}\)\(^{x^2}\)y+\(^{xy^2}\)−xy+\(\frac{1}{2}\)\(^{xy^2}\)−5xy−\(\frac{1}{2}\)\(^{xy^2}\)
P=(\(\frac{1}{3}\)−\(\frac{1}{3}\))+(\(^{xy^2}\)+\(\frac{1}{2}\)\(^{xy^2}\))+(−xy−5xy)
P=0+\(\frac{3}{2}\)\(^{xy^2}\)+(−6xy)=\(\frac{3}{2}\)\(^{xy^2}\)−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=\(\frac{-9}{4}\)=-2,25
