e, ĐK: \(x\ne2\)

\(\dfrac{3}{x-2}>1\Leftrightarrow\dfrac{5-x}{x-2}>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}5-x>0\\x-2>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}5-x< 0\\x-2< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow2< x< 5\)

\(\left(2\right)\Leftrightarrow\) vô nghiệm

Vậy \(2< x< 5\)

f, ĐK: \(x\ne\dfrac{1}{2}\)

\(\dfrac{2x^2+x}{1-2x}\ge1-x\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x-1\ge0\\\left(1-2x\right)\left(x-1\right)>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}4x-1\le0\\\left(1-2x\right)\left(x-1\right)< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{4}\\\dfrac{1}{2}< x< 1\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 1\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\left[{}\begin{matrix}x>1\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{4}\)

Vậy ...

\(\dfrac{x+1}{x+2}-\dfrac{5}{x-2}=\dfrac{20}{4-x^2}\) (\(ĐK:x\)≠\(2;-2\))

⇔ \(\dfrac{\left(x+1\right)\left(x-2\right)-5\left(x+2\right)}{x^2-4}=\dfrac{20}{4-x^2}\)

⇔ \(-\left(x+1\right)\left(x-2\right)+5\left(x+2\right)=20\)

⇔ \(-\left(x^2-2x+x-2\right)+5x+10=20\)

⇔ \(-x^2+x+2+5x+10-20=0\)

⇔ \(-x^2+6x-8=0\)

⇔ \(-\left(x^2-6x+9\right)=-1\)

⇔ \(\left(x-3\right)^2=1\)

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

⇔ \(\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Vậy ...

b: \(\Leftrightarrow20-5\left(3x+2\right)>4\left(x+7\right)\)

=>20-15x-10>4x+28

=>-15x+10-4x-28>0

=>-19x-18>0

=>-19x>18

hay x<-18/19

Đặt \(x^2+3x=a\left(a>=-\dfrac{9}{4}\right)\)

BPT sẽ trở thành \(a>=2+\sqrt{5a+14}\)

=>\(a-2>=\sqrt{5a+14}\)

=>\(\sqrt{5a+14}< =a-2\)

=>\(\left\{{}\begin{matrix}a-2>=0\\5a+14< =\left(a-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\5a+14-a^2+4a-4< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\-a^2+9a+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\a^2-9a-10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left(a-10\right)\left(a+1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left[{}\begin{matrix}a>=10\\a< =-1\end{matrix}\right.\end{matrix}\right.\)

=>a>=10

=>\(x^2+3x>=10\)

=>\(x^2+3x-10>=0\)

=>(x+5)(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =-5\end{matrix}\right.\)

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

* 2x - 1 = -(x + 2)

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....

=>x^2+4x+4-x^2-10x-25<=-8x-10

=>-6x-21<=-8x-10

=>2x<=11

=>x<=11/2

Ta có: \(-\dfrac{1}{x-2}\ge0\)

nên x-2<0

hay x<2

\(-\dfrac{1}{x-2}\ge0\Leftrightarrow x-2\le0\Leftrightarrow x\le2\)

Mà : $x ≠ 2 $

Do đó, bất phương trình vô nghiệm

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

ĐKXĐ: \(x-2\ne0\Leftrightarrow x\ne2\)

\(\Leftrightarrow-1\left(x-2\right)\ge0\)

\(\Leftrightarrow x-2\le0\)

\(\Leftrightarrow x\le2\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(không-thõa-mãn-ĐKXĐ\right)\\x< 2\left(thỏa-mãn-ĐKXĐ\right)\end{matrix}\right.\)

Vậy \(S=\left\{x|x< 2\right\}\)

\(\Leftrightarrow-x+2x>9-5\)

\(\Leftrightarrow x>4\)

`-x+5 > 9-2x`

`<=>-x+2x > 9-5`

`<=>x > 4`

Vậy `S={x|x > 4}`

-x+5 > 9-2x

\(\Leftrightarrow-x+5-9+2x>0\)

\(\Leftrightarrow x-4>0\Leftrightarrow x>4\)

\(x\left(x-1\right)>0\)

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