ĐKXĐ: \(x>=-\dfrac{1.96}{3}=-\dfrac{196}{300}=\dfrac{-49}{75}\)
\(\sqrt{\dfrac{1.96+3x}{4}}=\sqrt{0,04}+\dfrac{1}{4}\cdot\sqrt{\dfrac{256}{25}}\)
=>\(\sqrt{\dfrac{3x+1.96}{4}}=0.2+\dfrac{1}{4}\cdot\dfrac{16}{5}=1\)
=>\(\dfrac{3x+1,96}{4}=1\)
=>3x+1,96=4
=>3x=2,04
=>\(x=\dfrac{2.04}{3}=0.68\left(nhận\right)\)