Đáp án:
\(\begin{array}{l}
a)\,\,x = 5\\
b)\,\,x = 2\\
c)\,\,x = 4.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,\frac{{x + 1}}{3} = x - 3 \Leftrightarrow x + 1 = 3\left( {x - 3} \right)\\
\Leftrightarrow x + 1 = 3x - 9\\
\Leftrightarrow 2x = 10\\
\Leftrightarrow x = 5.\\
b)\,\,\frac{2}{x} - \frac{4}{{1 - x}} = \frac{{3x + 4}}{{{x^2} - x}}\\
DK:\,\,\,x \ne 0,\,\,\,x \ne 1\\
pt \Leftrightarrow \frac{2}{x} + \frac{4}{{x - 1}} = \frac{{3x + 4}}{{x\left( {x - 1} \right)}}\\
\Leftrightarrow 2\left( {x - 1} \right) + 4x = 3x + 4\\
\Leftrightarrow 2x - 2 + 4x = 3x + 4\\
\Leftrightarrow 3x = 6\\
\Leftrightarrow x = 2.\\
c)\,\,\,\sqrt {2x + 1} = 7 - x\\
DK:\,\,\,x \ge - \frac{1}{2}\\
pt \Leftrightarrow \left\{ \begin{array}{l}
7 - x \ge 0\\
2x + 1 = {\left( {7 - x} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 7\\
2x + 1 = 49 - 14x + {x^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 7\\
{x^2} - 16x + 48 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 7\\
\left[ \begin{array}{l}
x = 4\\
x = 12
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 4.
\end{array}\)
