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

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

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

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

\(\Leftrightarrow x^2+2x-2=x-2\)

\(\Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(kTM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-1\)

Ta có: \(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)

\(\Leftrightarrow x\left(x+2\right)-2=x-2\)

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

\(\Leftrightarrow x\left(x+1\right)=0\)

hay x=-1

a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)

\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)

\(\Leftrightarrow3-x-2+7x-7=0\)

\(\Leftrightarrow6x-6=0\)

hay x=1(loại

b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)

\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)

Suy ra: \(-2-x^2+5x-4=x^2+x-6\)

\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)

\(\Leftrightarrow-2x^2+4x=0\)

\(\Leftrightarrow-2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)

\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)

\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow3-x-2+7x-7=0\)

\(\Rightarrow6x-6=0\)

\(\Rightarrow x=1\)

a: Ta có: \(2x+3>1-x\)

\(\Leftrightarrow3x>-2\)

hay \(x>-\dfrac{2}{3}\)

b: Ta có: \(15-2\left(x-3\right)< -2x+5\)

\(\Leftrightarrow15-2x+6+2x-5< 0\)

\(\Leftrightarrow16< 0\left(vôlý\right)\)

c: Ta có: \(\left(x+1\right)\left(x-3\right)\le\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-3x+x-3-x^2+x-4x+4\le0\)

\(\Leftrightarrow-5x\le-1\)

hay \(x\ge\dfrac{1}{5}\)

d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)

\(\Leftrightarrow8x+4-6+6x\ge12-3x\)

\(\Leftrightarrow14x+3x\ge12+2=14\)

\(\Leftrightarrow x\ge\dfrac{14}{17}\)

e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)

\(\Leftrightarrow6x+12+4x-8< 6x-9\)

\(\Leftrightarrow4x< -9+8-12=-13\)

hay \(x< -\dfrac{13}{4}\)

f: Ta có: \(\left(x+1\right)\left(x-2\right)-\left(2-x\right)\left(3-x\right)>0\)

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

\(\Leftrightarrow x^2-x-2-x^2+5x-6>0\)

\(\Leftrightarrow4x>8\)

hay x>2

g: Ta có: \(\left(2x-1\right)^2\le2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-4x+1-2x^2+4x-2\le0\)

\(\Leftrightarrow2x^2\le1\)

\(\Leftrightarrow x^2\le\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{\sqrt{2}}{2}\le x\le\dfrac{\sqrt{2}}{2}\)

b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\)

\(\Leftrightarrow-2x^2+6x-4=0\)

a=-2; b=6; c=-4

Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)