a) ( 1/3x – 8/3 ) . (2,5 + -7/5 ) = 0 b) ( -5/4x + 3,25 ) . (3/5 – (-5/2x)) = 0 c) ( x – 2/7 ) . ( x + 3/4 ) = 0 d) (2x + 1/5 ) . (-3/5x + 4/7 ) = 0

`a,` `(1/3x-8/3).(2,5x+{-7}/5)=0`

`⇒(1/3x-8/3).(5/2x-7/5)=0`

$⇒\left[ \begin{array}{l}\dfrac13x-\dfrac{8}{3}=0\\\dfrac52x+\dfrac{-7}5=0\end{array} \right.$

$⇒\left[ \begin{array}{l}\dfrac13x=\dfrac{8}{3}\\\dfrac52x=\dfrac75\end{array} \right.$

$⇒\left[ \begin{array}{l}x=8\\x=\dfrac{14}{25}\end{array} \right.$

`b,` `({-5}/4x+3,25).(3/5-{-5}/2x)=0`

`⇒({-5}/4x+13/4).(3/5-{-5}/2x)=0`

$⇒\left[ \begin{array}{l}\dfrac{-5}4x+\dfrac{13}{4}=0\\\dfrac35-\dfrac{-5}2x=0\end{array} \right.$

$⇒\left[ \begin{array}{l}\dfrac{-5}4x=\dfrac{-13}{4}\\\dfrac35=\dfrac{-5}2x\end{array} \right.$

$⇒\left[ \begin{array}{l}x=\dfrac{13}5\\x=\dfrac{-6}{25}\end{array} \right.$

`c,` `( x-2/7).(x+3/4)=0`

$⇒\left[ \begin{array}{l}x-\dfrac27=0\\x+\dfrac34=\end{array} \right.$

$⇒\left[ \begin{array}{l}x=\dfrac27\\x=\dfrac{-3}4\end{array} \right.$

`d,` `(2x+1/5).(-3/5x+4/7)=0`

$⇒\left[ \begin{array}{l}2x+\dfrac15=0\\\dfrac{-3}5x+\dfrac47=0\end{array} \right.$

$⇒\left[ \begin{array}{l}2x=\dfrac{-1}5\\\dfrac{-3}5=\dfrac{-4}7\end{array} \right.$

$⇒\left[ \begin{array}{l}x=\dfrac{-1}{10}\\x=\dfrac{20}{21}\end{array} \right.$