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se
x=y=-1
vamos substituir x e y por (-1) e cacular as potências
![(2x^3y^2-x^2y)^5= \\ \\ .[2(-1)^3(-1)^2-(-1)^2(-1)]^5= \\ \\ .[2(-1)(+1)-(+1)(-1)]^5= \\ \\ .[2(-1)-(-1)]^5= \\ \\ (-2+1)^5= \\ \\ (-1)^5=\fbox{$-1$}](https://tex.z-dn.net/?f=%282x%5E3y%5E2-x%5E2y%29%5E5%3D+%5C%5C++%5C%5C+.%5B2%28-1%29%5E3%28-1%29%5E2-%28-1%29%5E2%28-1%29%5D%5E5%3D+%5C%5C++%5C%5C+.%5B2%28-1%29%28%2B1%29-%28%2B1%29%28-1%29%5D%5E5%3D+%5C%5C++%5C%5C+.%5B2%28-1%29-%28-1%29%5D%5E5%3D+%5C%5C++%5C%5C+%28-2%2B1%29%5E5%3D+%5C%5C++%5C%5C+%28-1%29%5E5%3D%5Cfbox%7B%24-1%24%7D "(2x^3y^2-x^2y)^5= \\ \\ .[2(-1)^3(-1)^2-(-1)^2(-1)]^5= \\ \\ .[2(-1)(+1)-(+1)(-1)]^5= \\ \\ .[2(-1)-(-1)]^5= \\ \\ (-2+1)^5= \\ \\ (-1)^5=\fbox{$-1$}")
x=y=-1
vamos substituir x e y por (-1) e cacular as potências
lavinnea:
blz!!♥
blz❤
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