a, \(y=f\left(x\right)=4x+1-\sqrt{3}\left(2x+1\right)=\left(4-2\sqrt{3}\right)x+1-\sqrt{3}\)
Lấy \(x_1;x_2\in R\left(x_1>x_2\right)\)
\(\Rightarrow y_1-y_2=\left(4-2\sqrt{3}\right)x_1-\left(4-2\sqrt{3}\right)x_2\)
\(=\left(4-2\sqrt{3}\right)\left(x_1-x_2\right)=\left(\sqrt{3}-1\right)^2\left(x_1-x_2\right)>0\Rightarrow y_1>y_2\)
Vậy hàm số đồng biến trên R
b, \(f\left(x\right)=\left(\sqrt{3}-1\right)^2x+1-\sqrt{3}=0\Leftrightarrow x=\frac{\sqrt{3}+1}{2}\)