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a: ĐKXĐ: x∈R
Lấy x1,x2 sao cho x1<x2
=>x1-x2<0
\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2+2x_1+3}-\sqrt{x_2^2+2x_2+3}}{x_1-x_2}\)
\(=\frac{x_1^2+2x_1+3-x_2^2-2x_2-3}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+2x_1+3}+\sqrt{x_2^2+2x_2+3}\right)}\)
\(=\frac{x_1^2-x_2^2+2x_1-2x_2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+2x_1+3}+\sqrt{x_2^2+2x_2+3}\right)}\)
\(=\frac{\left(x_1-x_2)\left(x_1+x_2\right)+2\left(x_1-2x_2\right)\right.}{\left(x_1-x_2\right)\left(\sqrt{x_1^2+2x_1+3}+\sqrt{x_2^2+2x_2+3}\right)}=\frac{x_1+x_2+2}{\sqrt{x_1^2+2x_1+3}+x_2^2+2x_2+3}\) >0
=>Hàm số đồng biến trên R
b: ĐKXĐ: \(x^2-3x+2\ge0\)
=>(x-1)(x-2)>=0
=>x>=2 hoặc x<=1
Lấy x1,x2 thuộc [2;+∞) sao cho 2<x1<x2
=>x1-2>0; x2-2>0
=>x1+x2-4>0
\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2-3x_1+2}-\sqrt{x_2^2-3x_2+2}}{x_1-x_2}\)
\(=\frac{x_1^2-3x_1+2-x_2^2+3x_2-2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{x_1^2-x_2^2-3x_1+3x_2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{\left(x_1^2-x_2^2\right)-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{\left(x_1-x_2\right)\left(x_1+x_2\right)_{}-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{\left(x_1+x_2\right)_{}-3}{\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}>0\)
=>Hàm số đồng biến trên khoảng [2;+∞)
Lấy x1,x2 thuộc (-∞;1] sao cho x2<x1<1
=>x1-1<0; x2-1<0
=>x1+x2-2<0
=>x1+x2-3<-1<0
\(\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\frac{\sqrt{x_1^2-3x_1+2}-\sqrt{x_2^2-3x_2+2}}{x_1-x_2}\)
\(=\frac{x_1^2-3x_1+2-x_2^2+3x_2-2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{x_1^2-x_2^2-3x_1+3x_2}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{\left(x_1^2-x_2^2\right)-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{\left(x_1-x_2\right)\left(x_1+x_2\right)_{}-3\left(x_1-x_2\right)}{\left(x_1-x_2\right)\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}\)
\(=\frac{\left(x_1+x_2\right)_{}-3}{\left(\sqrt{x_1^2-3x_1+2}+\sqrt{x_2^2-3x_2+2}\right)}>0\)
=>Hàm số nghịch biến trên khoảng (-∞;1]
\(DK:\hept{\begin{cases}-1\le x\le1\\x\ne0\end{cases}}\)
Ta co:
\(f\left(-x\right)=\frac{\sqrt{1-\left(-x\right)}+\sqrt{-x+1}}{\sqrt{-x+2}-\sqrt{2-\left(-x\right)}}=-\left(\frac{\sqrt{1-x}+\sqrt{x+1}}{\sqrt{x+2}-\sqrt{2-x}}\right)=-f\left(x\right)\)
Suy ra: f(x) la ham so chan