Verifique se são iguais as funções

\[\begin{array}{*{20}{l}}   {f\left( x \right) = \frac{{2 – x}}{{{x^2} – 4}}}&{\text{e}}&{g\left( x \right) = \frac{{ – 1}}{{x + 2}}} \end{array}\]

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\[\begin{array}{*{20}{l}}
{{D_f} = \left\{ {x \in \mathbb{R}:{x^2} – 4 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – 2,2} \right\}}&{\text{e}}&{{D_g} = \left\{ {x \in \mathbb{R}:x + 2 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – 2} \right\}}
\end{array}\]
As funções não são iguais, pois ${D_f} \ne {D_g}$.
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\[\begin{array}{*{20}{l}}   {f\left( x \right) = \frac{x}{{x – 1}}}&{\text{e}}&{g\left( x \right) = \frac{{{x^2} – x}}{{{{\left( {x – 1} \right)}^2}}}} \end{array}\]

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\[\begin{array}{*{20}{l}}
{{D_f} = \left\{ {x \in \mathbb{R}:x – 1 \ne 0} \right\} = \mathbb{R}\backslash \left\{ 1 \right\}}&{\text{e}}&{{D_g} = \left\{ {x \in \mathbb{R}:{{\left( {x – 1} \right)}^2} \ne 0} \right\} = \mathbb{R}\backslash \left\{ 1 \right\}}
\end{array}\]
\[g\left( x \right) = \frac{{{x^2} – x}}{{{{\left( {x – 1} \right)}^2}}} = \frac{{x\left( {x – 1} \right)}}{{{{\left( {x – 1} \right)}^2}}} = \frac{x}{{x – 1}} = f\left( x \right),\,\forall x \in {D_f}\]
As funções são iguais, pois ${D_f} = {D_g}$ e $f\left( x \right) = g\left( x \right),\,\forall x \in {D_f}$.
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\[\begin{array}{*{20}{l}}   {f\left( x \right) = \frac{{{x^4} – 25}}{{{x^2} + 5}}}&{\text{e}}&{g\left( x \right) = {x^2} – 5} \end{array}\]

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\[\begin{array}{*{20}{l}}
{{D_f} = \left\{ {x \in \mathbb{R}:{x^2} + 5 \ne 0} \right\} = \mathbb{R}}&{\text{e}}&{{D_g} = \left\{ {x \in \mathbb{R}:\left( {{x^2} – 5} \right) \in \mathbb{R}} \right\} = \mathbb{R}}
\end{array}\]
\[f\left( x \right) = \frac{{{x^4} – 25}}{{{x^2} + 5}} = \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} – 5} \right)}}{{{x^2} + 5}} = {x^2} – 5 = g\left( x \right),\,\forall x \in {D_f}\]
As funções são iguais, pois ${D_f} = {D_g}$ e $f\left( x \right) = g\left( x \right),\,\forall x \in {D_f}$.