Question

Me ajudem a resolver.
Determinar a resposta.
Não pode ser só resposta, tem que ter o cálculo. obrigada

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Fatorac_{\!\!,}\tilde ao\,do\,cubo\,perfeito}\\\sf u^3+3u^2v+3uv^2+v^3=(u+v)^3\\\underline{\rm Fatorac_{\!\!,}\tilde ao\,do\,trin\hat omio}\\\underline{\rm\,quadrado\,perfeito}\\\sf u^2+2uv+v^2=(u+v)^2\\\underline{\rm Fatorac_{\!\!,}\tilde ao\,da\,diferenc_{\!\!,}a\,de\,dois\,quadrados}\\\sf u^2-v^2=(u+v)(u-v)\\\underline{\rm Fatorac_{\!\!,}\tilde ao\,por\,fator}\\\underline{\rm\,comum\,em\,evid\hat encia}\\\sf k~\!u+kv=k\cdot(u+v)\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\bf Simplifiquemos\,primeiramente}\\\underline{\bf o\,polin\hat omio:}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\dfrac{x^3+3x^2y+3xy^2+y^3}{(zx^2+y^2z+2xyz)(x^2-y^2)}\\\\\sf\dfrac{(x+y)^3}{z(x^2+y^2+2xy)(x^2-y^2)}\\\\\sf\dfrac{(x+y)^3}{z(x+y)^2\cdot(x-y)\cdot(x+y)}\\\\\sf\dfrac{\diagdown\!\!\!\!(x+\diagdown\!\!\!\!y)^3}{z\diagdown\!\!\!\!\!(x+\diagdown\!\!\!\!y)^3\cdot( x-y)}\\\\\sf\dfrac{1}{z(x-y)}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Vamos\,substituir\,os\,valores:}\\\sf\dfrac{1}{z(x-y)}=\dfrac{1}{2011\cdot(2009-2010)}\\\\\sf\dfrac{1}{z(x-y)}=\dfrac{1}{2011(-1)}=-\dfrac{1}{2011}\\\\\Huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~e}}}}\end{array}}$