(\(\frac{-2}{3}\)x\(^3\)y\(^2\))(\(\frac{1}{2}\)x\(^2\)y\(^5\))
\(P=\left(\frac{-2}{3}x^3y^2\right)\left(\frac{1}{2}x^2y^5\right)\)
\(P=\frac{-2}{3}x^3y^2.\frac{1}{2}x^2y^5\)
\(P=\left(\frac{-2}{3}.\frac{1}{2}\right)x^{3+2}y^{2+5}\)
\(P=\frac{-1}{3}x^5y^7\)
Thay \(x=-1\); \(y=1\)vào P ta có:
\(P=\frac{-1}{3}\left(-1^5\right)1^7\)
\(P=\frac{-1}{3}.\left(-1\right).1\)
\(P=\frac{1}{3}.1=\frac{1}{3}\)
Vậy : \(P=\frac{1}{3}\)Tại \(x=-1\)\(y=1\)
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