\(\left(4x-3\right)\left(\dfrac{3}{5}x+\dfrac{1}{2}\right)=0\)
\(=>4x-3=0\) hoặc \(\dfrac{3}{5}x+\dfrac{1}{2}=0\)
\(=>x=\dfrac{3}{4}\) hoặc x = -5/6
(4x - 3).(\(\dfrac{3}{5}\)x + \(\dfrac{1}{2}\)) = 0
=> TH1: 4x - 3 = 0
=> 4x =3
=> x = loại
=>TH2: (\(\dfrac{3}{5}\)x + \(\dfrac{1}{2}\)) = 0
=> \(\dfrac{3}{5}\) x = \(\dfrac{1}{2}\)
=> x = \(\dfrac{1}{2}\): \(\dfrac{3}{5}\)
=> x = \(\dfrac{5}{6}\)
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