a: Lấy x1,x2 thuộc (1;+∞) sao cho 1<x1<x2
=>x1-1>0; x2-1>0
\(f\left(x\right)=\frac{2x^2-x-3}{x-1}\)
\(=\frac{2x^2-2x+x-1-2}{x-1}=2x+1-\frac{2}{x-1}\)
\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}\)
\(=\frac{2x_1+1-\frac{2}{x_1-1}-2x_2-1+\frac{2}{x_2-1}}{x_1-x_2}=\frac{2\left(x_1-x_2\right)-\frac{2}{x_1-1}+\frac{2}{x_2-1}}{x_1-x_2}\)
\(=\frac{2\left(x_1-x_2\right)+\frac{2\left(x_1-1\right)-2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}}{x_1-x_2}=\frac{2\left(x_1-x_2\right)+\frac{2\left(x_1-x_2\right)}{\left(x_1-1\right)\left(x_2-1\right)}}{x_1-x_2}\)
\(=2+\frac{2}{\left(x_1-1\right)\left(x_2-1\right)}>0\)
=>Hàm số đồng biến trên (1;+∞)
d: Lấy x1,x2 thuộc (-∞;0) sao cho x1<x2<0
\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2+1}-\sqrt{x_2^2+1}}{x_1-x_2}\)
\(=\frac{x_1^2+1-x_2^2-1}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+1}+\sqrt{x_2^2+1}\right)}=\frac{x_1+x_2}{\left(\sqrt{x_1^2+1}+\sqrt{x_2^2+1}\right)}<0\)
=>Hàm số nghịch biến trên (-∞;0)
