a) Đặt A(x)=0
\(\Leftrightarrow\dfrac{1}{6}-x^2=0\)
\(\Leftrightarrow x^2=\dfrac{1}{6}\)
hay \(x\in\left\{\dfrac{\sqrt{6}}{6};-\dfrac{\sqrt{6}}{6}\right\}\)
b) Đặt B(x)=0
\(\Leftrightarrow\left(9x-18\right)-\left(x-3^2\right)=0\)
\(\Leftrightarrow9x-18-x+9=0\)
\(\Leftrightarrow8x-9=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\)