Đáp án:
\(\frac{{3 - \sqrt {29} }}{2} < x < \frac{{3 + \sqrt {29} }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
5\left( {x - 1} \right) - x\left( {7 - x} \right) < x\\
\to 5x - 5 - 7x + {x^2} - x < 0\\
\to {x^2} - 3x - 5 < 0\left( 1 \right)\\
Xét:{x^2} - 3x - 5 = 0\\
Δ= 9 - 4.\left( { - 5} \right) = 29 > 0\\
\to \left[ \begin{array}{l}
x = \frac{{3 + \sqrt {29} }}{2}\\
x = \frac{{3 - \sqrt {29} }}{2}
\end{array} \right.\\
\left( 1 \right) \to \left( {x - \frac{{3 + \sqrt {29} }}{2}} \right)\left( {x - \frac{{3 - \sqrt {29} }}{2}} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - \frac{{3 + \sqrt {29} }}{2} > 0\\
x - \frac{{3 - \sqrt {29} }}{2} < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - \frac{{3 + \sqrt {29} }}{2} < 0\\
x - \frac{{3 - \sqrt {29} }}{2} > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > \frac{{3 + \sqrt {29} }}{2}\\
x < \frac{{3 - \sqrt {29} }}{2}
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x < \frac{{3 + \sqrt {29} }}{2}\\
x > \frac{{3 - \sqrt {29} }}{2}
\end{array} \right.
\end{array} \right.\\
KL:\frac{{3 - \sqrt {29} }}{2} < x < \frac{{3 + \sqrt {29} }}{2}
\end{array}\)
