Resolva as inequações
Funções racionais: Aleph 11 - Volume 2 Pág. 39 Ex. 13
Enunciado
Resolva, em $\mathbb{R}$, as equações:
- $5 + \frac{1}{x} > \frac{{16}}{x}$
- $1 + \frac{5}{{x – 1}} \leqslant \frac{7}{6}$
- $\frac{{{x^2} – 16}}{{{x^2} – 4x + 5}} \geqslant 0$
Resolução
- Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{5 + \frac{1}{x} > \frac{{16}}{x}}& \Leftrightarrow &{\frac{{5x + 1 – 16}}{x} > 0} \\
{}& \Leftrightarrow &{\frac{{5x – 15}}{x} > 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{l}}
{5x – 15 < 0} \\
{x < 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{5x – 15 > 0} \\
{x > 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{l}}
{x < 3} \\
{x < 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x > 3} \\
{x > 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{x < 0}& \vee &{x > 3}
\end{array}} \\
{}& \Leftrightarrow &{x \in \left] { – \infty ,0} \right[ \cup \left] {3, + \infty } \right[}
\end{array}\]Em alternativa, podemos usar um quadro de sinal (a partir da expressão obtida na 2.ª equivalência acima):
$x$ ${ – \infty }$ $0$ $3$ ${ + \infty }$ ${5x – 15}$ $ – $ $ – $ $ – $ $0$ $ + $ $x$ $ – $ $0$ $ + $ $ + $ $ + $ ${\frac{{5x – 15}}{x}}$ $ + $ n.d. $ – $ $0$ $ + $ Portanto, $\begin{array}{*{20}{l}}
{5 + \frac{1}{x} > \frac{{16}}{x}}& \Leftrightarrow &{x \in \left] { – \infty ,0} \right[ \cup \left] {3, + \infty } \right[}
\end{array}$.
- Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{1 + \frac{5}{{x – 1}} \leqslant \frac{7}{6}}& \Leftrightarrow &{\frac{{6\left( {x – 1} \right) + 6 \times 5 – 7 \times \left( {x – 1} \right)}}{{6\left( {x – 1} \right)}} \leqslant 0} \\
{}& \Leftrightarrow &{\frac{{6x – 6 + 30 – 7x + 7}}{{6\left( {x – 1} \right)}} \leqslant 0} \\
{}& \Leftrightarrow &{\frac{{31 – x}}{{6\left( {x – 1} \right)}} \leqslant 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{l}}
{31 – x \leqslant 0} \\
{6\left( {x – 1} \right) > 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{31 – x \geqslant 0} \\
{6\left( {x – 1} \right) < 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{l}}
{x \geqslant 31} \\
{x > 1}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x \leqslant 31} \\
{x < 1}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{x \geqslant 31}& \vee &{x < 1}
\end{array}} \\
{}& \Leftrightarrow &{x \in \left] { – \infty ,1} \right[ \cup \left[ {31, + \infty } \right]}
\end{array}\]Em alternativa, podemos usar um quadro de sinal (a partir da expressão obtida na 3.ª equivalência acima):
$x$ ${ – \infty }$ $1$ $31$ ${ + \infty }$ ${31 – x}$ $ + $ $ + $ $ + $ $0$ $ – $ ${6\left( {x – 1} \right)}$ $ – $ $0$ $ + $ $ + $ $ + $ ${\frac{{31 – x}}{{6\left( {x – 1} \right)}}}$ $ – $ n.d. $ + $ $0$ $ – $ Portanto, $\begin{array}{*{20}{l}}
{1 + \frac{5}{{x – 1}} \leqslant \frac{7}{6}}& \Leftrightarrow &{x \in \left] { – \infty ,1} \right[ \cup \left[ {31, + \infty } \right]}
\end{array}$.
- Tem-se sucessivamente:
\[\begin{array}{*{20}{l}}
{\frac{{{x^2} – 16}}{{{x^2} – 4x + 5}} \geqslant 0}& \Leftrightarrow &{\frac{{\left( {x + 4} \right)\left( {x – 4} \right)}}{{{x^2} – 4x + 5}} \geqslant 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{l}}
{\left( {x + 4} \right)\left( {x – 4} \right) \geqslant 0} \\
{{x^2} – 4x + 5 > 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{\left( {x + 4} \right)\left( {x – 4} \right) \leqslant 0} \\
{{x^2} – 4x + 5 < 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{\left\{ {\begin{array}{*{20}{l}}
{x \in \left] { – \infty , – 4} \right] \cup \left[ {4, + \infty } \right[} \\
{x \in \mathbb{R}}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x \in \left[ { – 4,4} \right]} \\
{x \in \emptyset }
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{x \in \left] { – \infty , – 4} \right] \cup \left[ {4, + \infty } \right[}
\end{array}\]Em alternativa, podemos usar um quadro de sinal:
$x$ ${ – \infty }$ ${ – 4}$ $4$ ${ + \infty }$ ${{x^2} – 16}$ $ + $ $0$ $ – $ $0$ $ + $ ${{x^2} – 4x + 5}$ $ + $ $ + $ $ + $ $ + $ $ + $ $\frac{{{x^2} – 16}}{{{x^2} – 4x + 5}}$ $ + $ $0$ $ – $ $0$ $ + $ Portanto, $\begin{array}{*{20}{l}}
{\frac{{{x^2} – 16}}{{{x^2} – 4x + 5}} \geqslant 0}& \Leftrightarrow &{x \in \left] { – \infty , – 4} \right] \cup \left[ {4, + \infty } \right[}
\end{array}$.
