Đáp án:
B1:
d) \(\left[ \begin{array}{l}
x = - 2\\
x = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)3x\left( {2x - 4} \right) - 7\left( {2x - 4} \right) = 0\\
\to \left( {2x - 4} \right)\left( {3x - 7} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{7}{3}
\end{array} \right.\\
b)\left( {x + 1} \right)\left( {2x + 3} \right) - \left( {3x - 2} \right)\left( {2x + 3} \right) = 0\\
\to \left( {2x + 3} \right)\left( {x + 1 - 3x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
- 2x + 3 = 0
\end{array} \right.\\
\to x = \dfrac{3}{2}\\
c)4x\left( {2x - 1} \right) - 4x + 2 = 0\\
\to 4x\left( {2x - 1} \right) - 2\left( {2x - 1} \right) = 0\\
\to \left( {2x - 1} \right)\left( {4x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
\to x = \dfrac{1}{2}\\
d)\left( {x - 2} \right)\left( {x + 2} \right) + 3\left( {x + 2} \right) = 0\\
\to \left( {x + 2} \right)\left( {x - 2 + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 2\\
x = - 1
\end{array} \right.\\
B2:\\
a)DK:x \ne \left\{ { - 3;2} \right\}\\
\dfrac{{x - 2 - 2\left( {x + 3} \right) - 3x - 5}}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = 0\\
\to - 2x - 7 - 2x - 6 = 0\\
\to 4x = - 13\\
\to x = - \dfrac{{13}}{4}\\
b)DK:x \ne \pm 1\\
\dfrac{{3\left( {x + 1} \right) + 4\left( {x - 1} \right) - x + 5}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to 3x + 3 + 4x - 4 - x + 5 = 0\\
\to 6x + 4 = 0\\
\to x = - \dfrac{2}{3}\\
d)DK:x \ne \pm 3\\
\dfrac{2}{{x - 3}} - \dfrac{3}{{x - 3}} = \dfrac{{7x + 5}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\to - \dfrac{1}{{x - 3}}\dfrac{{7x + 5}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\to \dfrac{{7x + 5 + x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\\
\to 8x + 8 = 0\\
\to x = - 1
\end{array}\)
