a: \(y=\frac{x^2+2}{x+1}\)

=>y'=\(\frac{\left(x^2+2\right)^{\prime}\left(x+1\right)-\left(x^2+2\right)\left(x+1\right)^{\prime}}{\left(x+1\right)^2}=\frac{2x\left(x+1\right)-\left(x^2+2\right)}{\left(x+1\right)^2}=\frac{2x^2+2x-x^2-2}{\left(x+1\right)^2}=\frac{x^2+2x-2}{\left(x+1\right)^2}\)

Đặt y'>0

=>\(x^2+2x-2>0\)

=>\(x^2+2x+1>3\)

=>\(\left(x+1\right)^2>3\)

=>\(\left[\begin{array}{l}x+1>\sqrt3\\ x+1<-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\sqrt3-1\\ x<-\sqrt3-1\end{array}\right.\)

Vậy: Hàm số đồng biến trên các khoảng \(\left(\sqrt3-1;+\infty\right);\left(-\infty;-\sqrt3-1\right)\)

Đặt y'<0

=>\(x^2+2x-2<0\)

=>\(x^2+2x+1<3\)

=>\(\left(x+1\right)^2<3\)

=>\(-\sqrt3

=>\(-\sqrt3-1

=>Hàm số nghịch biến trên khoảng \(\left(-\sqrt3-1;\sqrt3-1\right)\)

b: \(y=\frac{2x^2-3}{x-2}\)

=>y'=\(\frac{\left(2x^2-3\right)^{\prime}\cdot\left(x-2\right)-\left(2x^2-3\right)\cdot\left(x-2\right)^{\prime}}{\left(x-2\right)^2}=\frac{2\cdot2x\left(x-2\right)-\left(2x^2-3\right)}{\left(x-2\right)^2}=\frac{4x^2-8x-2x^2+3}{\left(x-2\right)^2}=\frac{2x^2-8x+3}{\left(x-2\right)^2}\)

Đặt y'>0

=>\(2x^2-8x+3>0\)

=>\(x^2-4x+1,5>0\)

=>\(x^2-4x+4>\frac52\)

=>\(\left(x-2\right)^2>\frac52=\frac{10}{4}\)

=>\(\left[\begin{array}{l}x-2>\frac{\sqrt{10}}{2}\\ x-2<-\frac{\sqrt{10}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{4+\sqrt{10}}{2}\\ x<\frac{4-\sqrt{10}}{2}\end{array}\right.\)

=>Hàm số đồng biến trên các khoảng \(\left(\frac{4+\sqrt{10}}{2};+\infty\right);\left(-\infty;\frac{4-\sqrt{10}}{2}\right)\)

Đặt y'<0

=>\(2x^2-8x+3<0\)

=>\(x^2-4x+1,5<0\)

=>\(x^2-4x+4<\frac52\)

=>\(\left(x-2\right)^2<\frac52=\frac{10}{4}\)

=>\(-\frac{\sqrt{10}}{2}

=>\(\frac{4-\sqrt{10}}{2}

=>Hàm số nghịch biến trên khoảng \(\left(\frac{4-\sqrt{10}}{2};\frac{4+\sqrt{10}}{2}\right)\)

c:

ĐKXĐ: x<>2; x<>-2

\(y=\frac{x+1}{x^2-4}\)

=>y'=\(\frac{\left(x+1\right)^{\prime}\cdot\left(x^2-4\right)-\left(x+1\right)\left(x^2-4\right)^{\prime}}{\left(x^2-4\right)^2}=\frac{x^2-4-2x\left(x+1\right)}{\left(x^2-4\right)^2}\)

\(=\frac{x^2-4-2x^2-2x}{\left(x^2-4\right)^2}=\frac{-x^2-2x-4}{\left(x^2-4\right)^2}=\frac{-x^2-2x-1-3}{\left(x^2-4\right)^2}=\frac{-\left(x+1\right)^2-3}{\left(x^2-4\right)^2}<0\forall x\) thỏa mãn ĐKXĐ

=>Hàm số luôn nghịch biến trên mọi khoảng ở TXĐ
d: ĐKXĐ: x<>1; x<>-1

\(y=\frac{2x+3}{x^2-1}\)

=>y'=\(\frac{\left(2x+3\right)^{\prime}\cdot\left(x^2-1\right)-\left(2x+3\right)\left(x^2-1\right)^{\prime}}{\left(x^2-1\right)^2}\)

\(=\frac{2\left(x^2-1\right)-2x\left(2x+3\right)}{\left(x^2-1\right)^2}=\frac{2x^2-2-4x^2-6x}{\left(x^2-1\right)^2}=\frac{-2x^2-6x-2}{\left(x^2-1\right)^2}\)

\(=\frac{-2\left(x^2+3x+1\right)}{\left(x^2-1\right)^2}\)

Đặt y'<0

=>\(-2\left(x^2+3x+1\right)<0\)

=>\(x^2+3x+1>0\)

=>\(x^2+3x+\frac94>\frac54\)

=>\(\left(x+\frac32\right)^2>\frac54\)

=>\(\left[\begin{array}{l}x+\frac32>\frac{\sqrt5}{2}\\ x+\frac32<-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{\sqrt5-3}{2}\\ x<\frac{-\sqrt5-3}{2}\end{array}\right.\)

=>Hàm số nghịch biến trên các khoảng \(\left(\frac{\sqrt5-3}{2};+\infty\right);\left(-\infty;\frac{-\sqrt5-3}{2}\right)\)

Đặt y'>0

=>\(-2\left(x^2+3x+1\right)>0\)

=>\(x^2+3x+1<0\)

=>\(x^2+3x+\frac94<\frac54\)

=>\(\left(x+\frac32\right)^2<\frac54\)

=>\(-\frac{\sqrt5}{2}

=>\(\frac{-\sqrt5-3}{2}

=>Hàm số đồng biến trên khoảng \(\left(\frac{-\sqrt5-3}{2};\frac{\sqrt5-3}{2}\right)\)