\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)

Ta có: \(\dfrac{x}{x+2}< \dfrac{x}{x+1}\)

\(\Leftrightarrow\dfrac{x}{x+2}-\dfrac{x}{x+1}< 0\)

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

\(\Leftrightarrow\dfrac{-x}{\left(x+2\right)\cdot\left(x+1\right)}< 0\)

Trường hợp 1: \(\left\{{}\begin{matrix}-x>0\\\left(x+2\right)\left(x+1\right)< 0\end{matrix}\right.\)

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

Trường hợp 2: \(\left\{{}\begin{matrix}-x< 0\\\left(x+2\right)\left(x+1\right)>0\end{matrix}\right.\)

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

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

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

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

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

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

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

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)

Biểu thức vế trái có nghĩa khi 

 \(x\ne-2;x\ne1\\ \dfrac{x-2}{x+2}+\dfrac{x+1}{x-1}\Leftrightarrow\dfrac{x-2}{x+2}+\dfrac{x+1}{x-1}-2>0\\ \dfrac{\left(x-2\right)\left(x-1\right)+\left(x+2\right)\left(x+1\right)-2\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\\ >0\\ \Leftrightarrow\dfrac{8-2x}{\left(x+2\right)\left(x-1\right)}>0\\ \Leftrightarrow\dfrac{4-x}{\left(x+2\right)\left(x-1\right)}>0\)  

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-4\right)< 0\) 

Lập bảng xét dấu

Vậy nghiệm của bất pt là 

\(x< -2.hay.1< x< 4\) 

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)

- TH1: Với \(x>1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

ĐKXĐ:\(x\ne-1\)

\(\dfrac{x-1}{x+1}\le5+x\\ \Leftrightarrow x-1\le\left(x+1\right)\left(5+x\right)\\ \Leftrightarrow x-1\le x^2+6x+5\\ \Leftrightarrow x^2+5x+6\ge0\\ \Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge-2\end{matrix}\right.\)