Sabendo que a razão
Funções racionais: Aleph 11 - Volume 2 Pág. 49 Ex. 8
Sabendo que a razão \[\frac{{x + 2}}{{x – 5}}\] é um valor maior do que $30$% de $x$, determine o valor de $x$.
\[\begin{array}{*{20}{l}}
{\frac{{x + 2}}{{x – 5}} > \frac{3}{{10}}x}& \Leftrightarrow &{\frac{{10x + 20 – 3{x^2} + 15x}}{{10\left( {x – 5} \right)}} > 0} \\
{}& \Leftrightarrow &{\frac{{ – 3{x^2} + 25x + 20}}{{10\left( {x – 5} \right)}} > 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{ – 3{x^2} + 25x + 20 > 0} \\
{10\left( {x – 5} \right) > 0}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{ – 3{x^2} + 25x + 20 < 0} \\
{10\left( {x – 5} \right) < 0}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x \in \left] {\frac{{ – 25 + \sqrt {865} }}{{ – 6}},\frac{{ – 25 – \sqrt {865} }}{{ – 6}}} \right[} \\
{x > 5}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x \in \left] { – \infty ,\frac{{ – 25 + \sqrt {865} }}{{ – 6}}} \right[ \cup \left] {\frac{{ – 25 – \sqrt {865} }}{{ – 6}}, + \infty } \right[} \\
{x < 5}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x \in \left] {\frac{{25 – \sqrt {865} }}{6},\frac{{25 + \sqrt {865} }}{6}} \right[} \\
{x > 5}
\end{array}} \right.}& \vee &{\left\{ {\begin{array}{*{20}{l}}
{x \in \left] { – \infty ,\frac{{25 – \sqrt {865} }}{6}} \right[ \cup \left] {\frac{{25 + \sqrt {865} }}{6}, + \infty } \right[} \\
{x < 5}
\end{array}} \right.}
\end{array}} \\
{}& \Leftrightarrow &{x \in \left] { – \infty ,\frac{{25 – \sqrt {865} }}{6}} \right[ \cup \left] {5,\frac{{25 + \sqrt {865} }}{6}} \right[}
\end{array}\]
