Ta có: \(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4\cdot\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)