Resolva, em $\mathbb{R}$, as equações
Funções racionais: Aleph 11 - Volume 2 Pág. 48 Ex. 1
Enunciado
Resolva, em $\mathbb{R}$, as equações:
- $a – \frac{5}{a} = 4$
- $\frac{9}{{x + 5}} = \frac{3}{{x – 3}}$
- $\frac{{x + 4}}{x} + \frac{3}{{x + 3}} = – \frac{{16}}{{{x^2} – 4x}}$
Resolução
- Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\mathop a\limits_{\left( a \right)} – \frac{5}{a} = \mathop 4\limits_{\left( a \right)} }& \Leftrightarrow &{\frac{{{a^2} – 5 – 4a}}{a} = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{{a^2} – 4a – 5 = 0}& \wedge &{a \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{a = \frac{{4 \pm \sqrt {16 + 20} }}{2}}& \wedge &{a \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left( {\begin{array}{*{20}{c}}
{a = – 1}& \vee &{a = 5}
\end{array}} \right)}& \wedge &{a \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{a = – 1}& \vee &{a = 5}
\end{array}}
\end{array}\] - Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\frac{9}{{\mathop {x + 5}\limits_{\left( {x – 3} \right)} }} = \frac{3}{{\mathop {x – 3}\limits_{\left( {x + 5} \right)} }}}& \Leftrightarrow &{\frac{{9x – 27 – 3x – 15}}{{\left( {x + 5} \right)\left( {x – 3} \right)}} = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{6x – 42 = 0}& \wedge &{\left( {x + 5} \right)\left( {x – 3} \right) \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{x = 7}& \wedge &{\left( {\begin{array}{*{20}{c}}
{x \ne – 5}& \wedge &{x \ne 3}
\end{array}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{x = 7}
\end{array}\] - Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\frac{{x + 4}}{{\mathop x\limits_{\left( {\left( {x + 3} \right)\left( {x – 4} \right)} \right)} }} + \frac{3}{{\mathop {x + 3}\limits_{\left( {x\left( {x – 4} \right)} \right)} }} = – \frac{{16}}{{\mathop {{x^2} – 4x}\limits_{\left( {x + 3} \right)} }}}& \Leftrightarrow &{\frac{{\left( {x + 3} \right)\left( {x – 4} \right)\left( {x + 4} \right) + 3x\left( {x – 4} \right) + 16\left( {x + 3} \right)}}{{x\left( {x + 3} \right)\left( {x – 4} \right)}} = 0} \\
{}& \Leftrightarrow &{\frac{{\left( {x + 3} \right)\left( {{x^2} – 16} \right) + 3x\left( {x – 4} \right) + 16\left( {x + 3} \right)}}{{x\left( {x + 3} \right)\left( {x – 4} \right)}} = 0} \\
{}& \Leftrightarrow &{\frac{{\left( {x + 3} \right){x^2} – 16\left( {x + 3} \right) + 3x\left( {x – 4} \right) + 16\left( {x + 3} \right)}}{{x\left( {x + 3} \right)\left( {x – 4} \right)}} = 0} \\
{}& \Leftrightarrow &{\frac{{\left( {x + 3} \right){x^2} + 3x\left( {x – 4} \right)}}{{x\left( {x + 3} \right)\left( {x – 4} \right)}} = 0} \\
{}& \Leftrightarrow &{\frac{{x\left[ {x\left( {x + 3} \right) + 3\left( {x – 4} \right)} \right]}}{{x\left( {x + 3} \right)\left( {x – 4} \right)}} = 0} \\
{}& \Leftrightarrow &{\frac{{x\left( {{x^2} + 6x – 12} \right)}}{{x\left( {x + 3} \right)\left( {x – 4} \right)}} = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{x\left( {{x^2} + 6x – 12} \right) = 0}& \wedge &{x\left( {x + 3} \right)\left( {x – 4} \right) \ne 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left( {\begin{array}{*{20}{c}}
{x = 0}& \vee &{{x^2} + 6x – 12 = 0}
\end{array}} \right)}& \wedge &{\left( {\begin{array}{*{20}{l}}
{x \ne 0}& \wedge &{x \ne – 3}& \wedge &{x \ne 4}
\end{array}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left( {\begin{array}{*{20}{c}}
{x = 0}& \vee &{x = \frac{{ – 6 \pm \sqrt {36 + 48} }}{2}}
\end{array}} \right)}& \wedge &{\left( {\begin{array}{*{20}{l}}
{x \ne 0}& \wedge &{x \ne – 3}& \wedge &{x \ne 4}
\end{array}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left( {\begin{array}{*{20}{l}}
{x = 0}& \vee &{x = – 3 – \sqrt {21} }& \vee &{x = }
\end{array} – 3 + \sqrt {21} } \right)}& \wedge &{\left( {\begin{array}{*{20}{l}}
{x \ne 0}& \wedge &{x \ne – 3}& \wedge &{x \ne 4}
\end{array}} \right)}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = – 3 – \sqrt {21} }& \vee &{x = – 3 + \sqrt {21} }
\end{array}}
\end{array}\]
