Đáp án:
$\begin{array}{l}
B1)a)5x - 10 = 0\\
\Rightarrow 5x = 10\\
\Rightarrow x = 2\\
Vậy\,x = 2\\
b)\left( {x - 2} \right)\left( {4x + 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 2 = 0\\
4x + 5 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{5}{4}
\end{array} \right.\\
Vậy\,x = - \dfrac{5}{4};x = 2\\
c)\dfrac{{4x + 5}}{3} + \dfrac{{2x - 1}}{2} = \dfrac{{3x - 4}}{6}\\
\Rightarrow \dfrac{{2\left( {4x + 5} \right) + 3\left( {2x - 1} \right)}}{6} = \dfrac{{3x - 4}}{6}\\
\Rightarrow 8x + 10 + 6x - 3 = 3x - 4\\
\Rightarrow 14x - 3x = - 4 + 3 - 10\\
\Rightarrow 9x = - 11\\
\Rightarrow x = - \dfrac{{11}}{9}\\
d)Dkxd:x \ne 1;x \ne - 1\\
\dfrac{x}{{x - 1}} - \dfrac{5}{{x + 1}} = \dfrac{{{x^2} + 1}}{{{x^2} - 1}}\\
\Rightarrow \dfrac{{x\left( {x + 1} \right) - 5\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{{x^2} + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\Rightarrow {x^2} + x - 5x + 5 = {x^2} + 1\\
\Rightarrow 4x = 4\\
\Rightarrow x = 1\left( {ktm} \right)\\
\text{Vậy pt vô nghiệm}
\end{array}$
$\begin{array}{l}
B2)a)x = 3\\
\Rightarrow m.3 + 3 = 2.3 + m\\
\Rightarrow 3m + 3 = m + 6\\
\Rightarrow 2m = 3\\
\Rightarrow m = \dfrac{3}{2}\\
b)mx + 3 = 2x + m\\
\Rightarrow \left( {m - 2} \right).x = m - 3\\
\Rightarrow \left\{ \begin{array}{l}
m - 2 = 0\\
m - 3 \ne 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m = 2\\
m \ne 3
\end{array} \right.\\
\Rightarrow m = 2
\end{array}$
Vậy m=2 thì pt vô nghiệm
