Đáp án:
a) x=2
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;1} \right\}\\
\dfrac{{x\left( {2x + 3} \right) - 7\left( {x - 1} \right) - 7}}{{x\left( {x - 1} \right)}} = 0\\
\to 2{x^2} + 3x - 7x + 7 - 7 = 0\\
\to 2{x^2} - 4x = 0\\
\to 2x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x = 2
\end{array} \right.\\
b)Đặt:{x^2} + 4x = t\\
Pt \to t\left( {t - 2} \right) = 15\\
\to {t^2} - 2t - 15 = 0\\
\to {t^2} - 5t + 3t - 15 = 0\\
\to t\left( {t - 5} \right) + 3\left( {t - 5} \right) = 0\\
\to \left( {t - 5} \right)\left( {t + 3} \right) = 0\\
\to \left[ \begin{array}{l}
t = 5\\
t = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 4x = 5\\
{x^2} + 4x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x + 5} \right)\left( {x - 1} \right) = 0\\
\left( {x + 1} \right)\left( {x + 3} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 5\\
x = 1\\
x = - 1\\
x = - 3
\end{array} \right.
\end{array}\)
