Lời giải:
Xét thấy \(a=0\Rightarrow \frac{-b^2}{b^2}=\frac{3}{4}\Leftrightarrow -1=\frac{3}{4}\) (vô lý)
\(b=0\Rightarrow \frac{3a^2}{a^2}=\frac{3}{4}\Leftrightarrow 3=\frac{3}{4}\) (vô lý)
Do đó \(a,b\neq 0\)
Khi đó, đặt \(a=tb\)
Ta có \(\frac{3}{4}=\frac{3a^2-b^2}{a^2+b^2}=\frac{3b^2t^2-b^2}{b^2t^2+b^2}=\frac{b^2(3t^2-1)}{b^2(t^2+1)}=\frac{3t^2-1}{t^2+1}\)
\(\Leftrightarrow 3(t^2+1)=4(3t^2-1)\Leftrightarrow t^2=\frac{7}{9}\)
\(\Rightarrow \frac{a}{b}=t=\pm \sqrt{\frac{7}{9}}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
=> \(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)= \(4.\left(3a^2+b^2\right)=3.\left(a^2+b^2\right)\)
=> \(12a^2+4b^2=3a^2+3b^2\)
=> \(12a^2+\left(-3a\right)^2=\left(-4b\right)^2+3b^2\)
=> \(9a^2=-1b^2\)
=> \(\left(\dfrac{a}{b}\right)^2=-\dfrac{1}{9}\)
=> \(\left(\dfrac{a}{b}\right)^2=\left(-\dfrac{1}{9}\right)^{ }\)
=> \(\dfrac{a}{b}=-\dfrac{1}{3}\)
Vậy:..........