Đáp án:
\(\begin{array}{l}
B3:\\
a)x = - 13\\
b)\left[ \begin{array}{l}
x = 2 + \sqrt {13} \\
x = 2 - \sqrt {13}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 2\\
x = 5
\end{array} \right.\\
B4:\\
a)x > - \dfrac{1}{2}\\
b)x \le \dfrac{6}{{11}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)5x - 15 + 3x + 15 = 7x - 13\\
\to 8x = 7x - 13\\
\to x = - 13\\
b)DK:x \ne \left\{ {0;3} \right\}\\
\dfrac{{2x - 4}}{{x - 3}} - \dfrac{6}{{x\left( {x - 3} \right)}} = \dfrac{{x - 1}}{x}\\
\to x\left( {2x - 4} \right) - 6 = \left( {x - 3} \right)\left( {x - 1} \right)\\
\to 2{x^2} - 8x - 6 = {x^2} - 4x + 3\\
\to {x^2} - 4x - 9 = 0\\
\to {x^2} - 4x + 4 = 13\\
\to {\left( {x - 2} \right)^2} = 13\\
\to \left[ \begin{array}{l}
x - 2 = \sqrt {13} \\
x - 2 = - \sqrt {13}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2 + \sqrt {13} \\
x = 2 - \sqrt {13}
\end{array} \right.\\
c)\left( {x - 2} \right)\left( {2x - 5} \right) = x\left( {x - 2} \right)\\
\to \left( {x - 2} \right)\left( {2x - 5} \right) - x\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {2x - 5 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 5
\end{array} \right.\\
B4:\\
a)4x + 8 - 1 > 2x + 6\\
\to 2x > - 1\\
\to x > - \dfrac{1}{2}\\
b)\dfrac{{14 - 7x - 12 - 2.2\left( {x - 1} \right)}}{6} \ge 0\\
\to 2 - 7x - 4x + 4 \ge 0\\
\to - 11x \ge - 6\\
\to x \le \dfrac{6}{{11}}
\end{array}\)
