Đáp án:
$a)\dfrac{1}{2}x + \dfrac{5}{2} = \dfrac{7}{2}x - \dfrac{3}{4}$
$⇔\dfrac{1}{2}x - \dfrac{7}{2}x = -\dfrac{3}{4} - \dfrac{5}{2}$
$⇔ -3x = -\dfrac{13}{4}$
$⇔ x = -\dfrac{13}{4} : (-3) $
$⇔ x = \dfrac{13}{12}$
$\text{Vậy x = $\dfrac{13}{12}$}$
$b)(x - \dfrac{2}{7})(x+ \dfrac{3}{4}) = 0$
⇔\(\left[ \begin{array}{l}x-\dfrac{2}{7}=0\\x+\dfrac{3}{4}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{2}{7}\\x=-\dfrac{3}{4}\end{array} \right.\)
$\text{Vậy x ∈ {$\dfrac{2}{7}$ ; -$\dfrac{3}{4}$ } }$
$c)(2x+\dfrac{1}{2})(-\dfrac{3}{5}x + \dfrac{4}{7}) = 0$
⇔\(\left[ \begin{array}{l}2x+\dfrac{1}{5}=0\\-\dfrac{3}{5}x+\dfrac{4}{7}=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{1}{10}\\x=\dfrac{20}{21}\end{array} \right.\)
$\text{Vậy x ∈ { -$\dfrac{1}{10}$ ; $\dfrac{20}{21}$}}$
$d)(-\dfrac{5}{4}x + 3,25)[ \dfrac{3}{5}- (-\dfrac{5}{2}x)]=0$
$⇔(-\dfrac{5}{4}x + 3,25)(\dfrac{3}{5} + \dfrac{5}{2}x)=0$
⇔\(\left[ \begin{array}{l}-\dfrac{5}{4}x +3,25=0\\\dfrac{3}{5}+\dfrac{5}{2}x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{13}{5}\\x=-\dfrac{6}{25}\end{array} \right.\)
$\text{Vậy x ∈ { $\dfrac{13}{5}$ ; -$\dfrac{6}{25}$}}$
