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[tex]\sf\dfrac{x+1}{x+2}=0\\\boxed{\begin{array}{l}\underline{\rm condic_{\!\!,}\tilde ao~de~exist\hat encia:}\\\sf x+2\ne0\\\sf x\ne-2\end{array}}\\\sf\dfrac{x+1}{x+2}=0\\\underline{\rm multiplicando~a~igualdade~por~x+2:}\\\sf \diagup\!\!\!\!(x+\diagup\!\!\!\!\!\!2)\cdot\dfrac{x+1}{\diagup\!\!\!\!\!(x+\diagup\!\!\!\!\!2)}=(x+2)\cdot0\\\sf x+1=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=-1}}}}\blue{\checkmark}[/tex]
