Resolva, em $\mathbb{R}$, as seguintes inequações
Funções racionais: Aleph 11 - Volume 2 Pág. 49 Ex. 6
Enunciado
Resolva, em $\mathbb{R}$, as seguintes inequações:
- \[\frac{{3x + 2}}{{x + 3}} > – \frac{2}{3}\]
- \[\frac{{x + 1}}{{x – 1}} – \frac{{x – 1}}{{x + 1}} > 0\]
- \[\frac{{a – 2}}{a} < \frac{{a – 4}}{{a – 6}}\]
Resolução
- Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\frac{{3x + 2}}{{\mathop {x + 3}\limits_{\left( 3 \right)} }} > – \frac{2}{{\mathop 3\limits_{\left( {x + 3} \right)} }}}& \Leftrightarrow &{\frac{{9x + 6 + 2x + 6}}{{3\left( {x + 3} \right)}} > 0} \\
{}& \Leftrightarrow &{\frac{{11x + 12}}{{3\left( {x + 3} \right)}} > 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{11x + 2 > 0} \\
{3\left( {x + 3} \right) > 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{11x + 2 < 0} \\
{3\left( {x + 3} \right) < 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x > – \frac{2}{{11}}} \\
{x > – 3}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x < – \frac{2}{{11}}} \\
{x < – 3}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x > – \frac{2}{{11}}}& \vee &{x < – 3}
\end{array}} \\
{}& \Leftrightarrow &{x \in \left] { – \infty , – 3} \right[ \cup \left] { – \frac{2}{{11}}, + \infty } \right[}
\end{array}\] - Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\frac{{x + 1}}{{\mathop {x – 1}\limits_{\left( {x + 1} \right)} }} – \frac{{x – 1}}{{\mathop {x + 1}\limits_{\left( {x – 1} \right)} }} > 0}& \Leftrightarrow &{\frac{{{x^2} + 2x + 1 – {x^2} + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} > 0} \\
{}& \Leftrightarrow &{\frac{{2x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} > 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{2x > 0} \\
{\left( {x – 1} \right)\left( {x + 1} \right) > 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{2x < 0} \\
{\left( {x – 1} \right)\left( {x + 1} \right) < 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x > 0} \\
{x \in \left] { – \infty , – 1} \right[ \cup \left] {1, + \infty } \right[}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x < 0} \\
{x \in \left] { – 1,1} \right[}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x \in \left] {1, + \infty } \right[}& \vee &{x \in \left] { – 1,0} \right[}
\end{array}} \\
{}& \Leftrightarrow &{x \in \left] { – 1,0} \right[ \cup \left] {1, + \infty } \right[}
\end{array}\]Em alternativa, podemos usar um quadro de sinal (a partir da expressão obtida na 2.ª equivalência acima):
$x$ ${ – \infty }$ $-1$ $0$ $1$ ${ + \infty }$ $2x$ $ – $ $ – $ $ – $ $0$ $ + $ $ + $ $ + $ ${\left( {x – 1} \right)\left( {x + 1} \right)}$ $ + $ $0$ $ – $ $ – $ $ – $ $0$ $ + $ ${\frac{{2x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}}$ $ – $ n.d. $ + $ $0$ $ – $ n.d. $ + $ Portanto, $\frac{{x + 1}}{{x – 1}} – \frac{{x – 1}}{{x + 1}} > 0 \Leftrightarrow x \in \left] { – 1,0} \right[ \cup \left] {1, + \infty } \right[$.
- Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\frac{{a – 2}}{{\mathop a\limits_{\left( {a – 6} \right)} }} < \frac{{a – 4}}{{\mathop {a – 6}\limits_{\left( a \right)} }}}& \Leftrightarrow &{\frac{{{a^2} – 8a + 12 – {a^2} + 4a}}{{a\left( {a – 6} \right)}} < 0} \\
{}& \Leftrightarrow &{\frac{{ – 4a + 12}}{{a\left( {a – 6} \right)}} < 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{ – 4a + 12 > 0} \\
{a\left( {a – 6} \right) < 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{ – 4a + 12 < 0} \\
{a\left( {a – 6} \right) > 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{a < 3} \\
{a \in \left] {0,6} \right[}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{a > 3} \\
{x \in \left] { – \infty ,0} \right[ \cup \left] {6, + \infty } \right[}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{a \in \left] {0,3} \right[}& \vee &{a \in \left] {6, + \infty } \right[}
\end{array}} \\
{}& \Leftrightarrow &{a \in \left] {0,3} \right[ \cup \left] {6, + \infty } \right[}
\end{array}\]Em alternativa, podemos usar um quadro de sinal (a partir da expressão obtida na 2.ª equivalência acima):
$a$ ${ – \infty }$ $0$ $3$ $6$ ${ + \infty }$ ${ – 4a + 12}$ $ + $ $ + $ $ + $ $0$ $ – $ $ – $ $ – $ ${a\left( {a – 6} \right)}$ $ + $ $0$ $ – $ $ – $ $ – $ $0$ $ + $ ${\frac{{ – 4a + 12}}{{a\left( {a – 6} \right)}}}$ $ + $ n.d. $ – $ $0$ $ + $ n.d. $ – $ Portanto, $\frac{{a – 2}}{a} < \frac{{a – 4}}{{a – 6}} \Leftrightarrow a \in \left] {0,3} \right[ \cup \left] {6, + \infty } \right[$.
