\(Giải:\)

\(ĐK:x\ne\left(-2\right);x\ne\left(-1\right)\)

\(\frac{x^2+2x+2}{x+1}>\frac{x^2+4x+5}{x+2}-1\Leftrightarrow\frac{x^2+2x+2}{x+1}>\frac{x^2+3x+3}{x+2}\)

\(\Leftrightarrow\frac{x^2+2x+1}{x+1}+\frac{1}{x+1}-\frac{x^2+3x+2+1}{x+2}>0\)

\(\Leftrightarrow\frac{\left(x+1\right)^2}{x+1}-\frac{\left(x+1\right)\left(x+2\right)}{x+2}+\frac{1}{x+1}-\frac{1}{x+2}>0\)

\(\Leftrightarrow x+1-x-1+\frac{1}{x+1}-\frac{1}{x+2}>0\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}>0\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}=\frac{1}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x+1>0\\x+2>0\end{cases}}hoặc\hept{\begin{cases}x+1< 0\\x+2< 0\end{cases}}\)

\(+,\hept{\begin{cases}x+1>0\\x+2>0\end{cases}}\Rightarrow x>\left(-2\right)\)

\(+,\hept{\begin{cases}x+1< 0\\x+2< 0\end{cases}}\Rightarrow x< \left(-2\right)\)

BPT đã được giải quyết

\(\Leftrightarrow\dfrac{1}{x-1}>\dfrac{1}{x-2}-\dfrac{1}{x+2}=\dfrac{\left(x+2\right)-\left(x-2\right)}{x^2-4}=\dfrac{4}{x^2-4}\)\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{4}{x^2-4}>0\Leftrightarrow\dfrac{x^2-4-4x+4}{\left(x-2\right)\left(x-1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x+2\right)}>0\)

Điều kiện tồn tại A

\(\left\{{}\begin{matrix}x\ne2\\x\ne1\\x\ne-2\end{matrix}\right.\) \(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+2\right)}\)

\(\left\{{}\begin{matrix}x>0\\\left[{}\begin{matrix}x< -2\\x>1\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x>1\)(1)

\(\left\{{}\begin{matrix}x< 0\\-2< x< 1\end{matrix}\right.\) \(\Rightarrow-2< x< 0\)(2)

từ (1)&(2)kết luận\(\Rightarrow\left[{}\begin{matrix}-2< x< 0\\x>1\end{matrix}\right.\)

ĐKXĐ: x<>0 và \(\begin{cases}x+\frac{1}{x^2}\ge0\\ x-\frac{1}{x^2}\ge0\end{cases}\)

=>\(\begin{cases}x^3+1\ge0\\ x^3-1\ge0\end{cases}\)

=>x>=1

\(\sqrt{x+\frac{1}{x^2}}+\sqrt{x-\frac{1}{x^2}}>\frac{2}{x}\)

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

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

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

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

=>\(\sqrt{x^2 - \frac{1}{x^4}}>\frac{2}{x^2}-x\)

TH1: \(\frac{2}{x^2}-x\le0\)

=>\(\frac{2 - x^3}{x^2}\le0\)

=>\(x^3\ge2\)

=>\(x\ge\sqrt[3]{2}\)

Khi \(x\ge\sqrt[3]{2}\) thì VT<=0; VT>0

=>Bất phương trình luôn đúng với \(x\ge\sqrt[3]{2}\) (2)

TH2: \(\frac{2}{x^2}-x>0\)

=>\(\frac{2 - x^3}{x^2}>0\)

=>\(x^3<2\)

=>\(x<\sqrt[3]{2}\)

BPT sẽ tương đương: \(x^2 - \frac{1}{x^4} > \left( \frac{2}{x^2} - x \right)^2\)

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

=>\(-\frac{1}{x^4} > \frac{4}{x^4} - \frac{4}{x}\)

=>\(\frac{4}{x} > \frac{5}{x^4}\)

=>\(4x^3>5\)

=>\(x^3>\frac54=\frac{10}{8}\)

=>\(x>\frac{\sqrt[3]{10}}{2}\)

=>\(\frac{\sqrt[3]{10}}{2} (1)

Từ (1),(2) suy ra \(S = \left( \sqrt[3]{\frac{5}{4}}; +\infty \right)\)

Ta có: \(\dfrac{1}{x^2} + \dfrac{x^2}{1-x^2} + \dfrac{5}{2}\left(\dfrac{\sqrt{1-x^2}}{x} + \dfrac{x}{\sqrt{1-x^2}}\right) + 2 > 0\) (1)

ĐKXĐ: \(\begin{cases}1-x^2>0\\ x<>0\end{cases}\Rightarrow\begin{cases}x^2<1\\ x<>0\end{cases}\)

=>-1<x<1 và x<>0

Đặt \(y=\dfrac{\sqrt{1-x^2}}{x}+\dfrac{x}{\sqrt{1-x^2}}\)

\(y^2 = \left(\dfrac{\sqrt{1-x^2}}{x}\right)^2 + 2 + \left(\dfrac{x}{\sqrt{1-x^2}}\right)^2 = \dfrac{1-x^2}{x^2} + 2 + \dfrac{x^2}{1-x^2}\)

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

=>\(\dfrac{1}{x^2} + \dfrac{x^2}{1-x^2} = y^2 - 1\)

Thay vào BPT ban đầu, ta được: \((y^2 - 1) + \dfrac{5}{2}y + 2 > 0\)

=>\(y^2+\dfrac{5}{2}y+1>0\iff2y^2+5y+2>0\)

=>(2y+1)(y+2)>0

=>y>-1/2 hoặc y<-2

TH1: 0<x<1

Theo AM-GM, ta được: \(y \ge 2 \cdot \sqrt{\dfrac{\sqrt{1-x^2}}{x} \cdot \dfrac{x}{\sqrt{1-x^2}}} = 2\)

Nếu y>-1/2 thì vì y>=2>-1/2

nên (1) luôn đúng

Nếu y<-2 thì vì y>=2>-2

nên (1) vô nghiệm

TH2: -1<x<0

=>\(\frac{\sqrt{1-x^2}}{x}<0;\frac{x}{\sqrt{1-x^2}}<0\)

Đặt \(t=-\dfrac{\sqrt{1-x^2}}{x}\)

=>\(y=-t-\dfrac{1}{t}=-\left(t+\dfrac{1}{t}\right)\le-2\)

Nếu y>-1/2 thì vì y<=-2<-1/2

nên (1) luôn sai

=>(1) vô nghiệm

Nếu y<-2 thì vì y<=-2<-2

nên mọi giá trị của x sẽ thỏa mãn y, trừ giá trị thỏa mãn y=-2

Đặt y=-2

=>\(t+\dfrac{1}{t}=2\iff t=1\)

=>\(-\dfrac{\sqrt{1-x^2}}{x}=1\)

\(\iff\sqrt{1-x^2}=-x\)

=>\(1-x^2=x^2\)

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

=>\(x=-\dfrac{1}{\sqrt{2}}\)

Do đó: x∈\(\left(-1;\frac{-1}{\sqrt2}\right)\) \(\cup\) \(\left(-\frac{1}{\sqrt2};0\right)\)

Vậy: \(S = \left(-1, -\dfrac{1}{\sqrt{2}}\right) \cup \left(-\dfrac{1}{\sqrt{2}}, 0\right) \cup (0, 1)\)

a: ĐKXĐ: x<0 hoặc x>1

TH1: x<0

=>x-1<-1<0

\(\sqrt{x^2 - x} = \sqrt{x(x-1)}\)

\(\frac{x - 1}{(x - 1) - \sqrt{-x(1-x)}} = \frac{-(1-x)}{-(1-x) - \sqrt{-x}\sqrt{1-x}} = \frac{\sqrt{1-x}}{\sqrt{1-x} + \sqrt{-x}}\)

Bất phương trình sẽ trở thành: \(\frac{\sqrt{1-x}}{\sqrt{1-x} + \sqrt{-x}} \ge 2x\)

x<0

=>2x<0

mà \(\frac{\sqrt{1-x}}{\sqrt{1-x} + \sqrt{-x}}>0\forall x\) thỏa mãn

nên Bất phương trình thỏa mãn với mọi x<0

TH2: x>1

=>x-1>0 và \(\sqrt{x^2 - x} = \sqrt{x(x-1)} = \sqrt{x}\sqrt{x-1}\)

Bất phương trình sẽ tương đương: \(\frac{\sqrt{x-1}}{\sqrt{x-1} - \sqrt{x}} \ge 2x\)

mà \(\frac{\sqrt{x-1}}{\sqrt{x-1} - \sqrt{x}}=\frac{\sqrt{x-1}(\sqrt{x-1} + \sqrt{x})}{(\sqrt{x-1} - \sqrt{x})(\sqrt{x-1} + \sqrt{x})}=\frac{\sqrt{x-1}(\sqrt{x-1} + \sqrt{x})}{(x-1) - x}=-\sqrt{x-1}(\sqrt{x-1}+\sqrt{x})\)

nên \(-\sqrt{x-1}(\sqrt{x-1} + \sqrt{x}) \ge 2x\)

=>\(\sqrt{x-1}(\sqrt{x-1}+\sqrt{x})+2x\le0\) (vô lý vì \(\sqrt{x-1}(\sqrt{x-1}+\sqrt{x})+2x>0\) với mọi x>1)

=>x∈∅

Vậy: x<0

ĐKXĐ: \(x\ne1,-1\)

Ta có: \(\dfrac{x-2}{x+1}\ge\dfrac{3x+2}{x-1}-2\)

\(\dfrac{x-2}{x+1}\ge\dfrac{3x+2-2\left(x-1\right)}{x-1}\)

\(\dfrac{x-2}{x+1}-\dfrac{3x+2-2x+2}{x-1}\ge0\)

\(\dfrac{x-2}{x+1}-\dfrac{x+4}{x-1}\ge0\)

\(\dfrac{\left(x-2\right)\left(x-1\right)-\left(x-4\right)\left(x+1\right)}{x^2-1}\ge0\)

\(\dfrac{x^2-3x+2-x^2+3x+4}{x^2-1}\ge0\)

\(\dfrac{6}{x^2-1}\ge0\)

\(\Rightarrow x^2-1>0\Leftrightarrow x^2>1\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\)(TM)

\(BPT\Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\ge\dfrac{\left(3x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow x^2-x-2x+2-3x^2-3x-2x-2-2x^2-2\ge0\)

\(\Leftrightarrow-4x^2-8x-2\ge0\)

\(\Leftrightarrow x^2+2x+\dfrac{1}{2}\ge0\)

\(\Leftrightarrow\left(x+1\right)^2-\dfrac{1}{2}\ge0\)

Vậy bất phương trình luôn đúng \(\forall x\).

\(\dfrac{15x-2}{4}-\dfrac{x^2+1}{3}>\dfrac{x\left(1-2x\right)}{6}+\dfrac{x-3}{2}\\ \Leftrightarrow3\left(15x-2\right)-4\left(x^2+1\right)>2x\left(1-2x\right)+6\left(x-3\right)\\ \Leftrightarrow45x-6-4x^2-4>2x-4x^2+6x-18\\ \Leftrightarrow45x-6x-2x>6+4-18\\ \Leftrightarrow37x>-8\\ \Leftrightarrow x>-\dfrac{8}{37}\)

\(\dfrac{3\left(15x-2\right)}{12}-\dfrac{4\left(x^2+1\right)}{12}>\dfrac{2x\left(1-2x\right)}{12}+\dfrac{6\left(x-3\right)}{12}\)

\(45x-6-\left(4x^2+4\right)>2x-4x^2+6x-18\)

\(45x-4x^2+4x^2-2x-6x>6+4-18\)

\(37x>-8\)

\(x>\dfrac{-8}{37}\)