\(\Leftrightarrow\dfrac{x}{a}-x>4-a-\dfrac{3}{a}\)

\(\Leftrightarrow x\left(\dfrac{1}{a}-1\right)>\dfrac{4a-a^2-3}{a}\)

- Nếu \(\dfrac{1}{a}-1>0\Leftrightarrow0< a< 1\)

\(\Rightarrow x>\dfrac{4a-a^2-3}{a\left(\dfrac{1}{a}-1\right)}\Leftrightarrow x>a-3\)

- Nếu \(\dfrac{1}{a}-1< 0\Leftrightarrow\left[{}\begin{matrix}a< 0\\a>1\end{matrix}\right.\)

\(\Rightarrow x< \dfrac{4a-a^2-3}{a\left(\dfrac{1}{a}-a\right)}\Leftrightarrow x< a-3\)

2x+4<a² -ax

2x-ax<a² -4

(2-a)x<(a--2)(a+2)

-(2-a)x >(2-a)(2+a)

-x>2+a

=> x<-(2+a)

\(\Leftrightarrow2x+4+ax-a^2<0\)

\(\Leftrightarrow\left(2+a\right)x<\left(a-2\right)\left(a+2\right)\)

nếu a=-2=> vô nghiệm

nếu a<-2=>x>(a-2)

nếu a>-2=> x<(a-2)

ĐKXĐ: x\(\ne3,x\ne-3\) 

\(\Rightarrow\left(x-a\right)\left(a-3\right)+\left(x+3\right)\left(a+3\right)=-6a\) 

\(\Leftrightarrow xa-3x-a^2+3a+ax+3x+3a+3=-6a\)

\(\Leftrightarrow2ax-a^2+12a+3=0\) \(\Leftrightarrow2ax=a^2-12a-3\Leftrightarrow x=\dfrac{a^2}{2}-6a-\dfrac{3}{2}\)(TM)

Vậy...

bạn lm sai rồix-3 chứ ko phải x+3 từ dòng thứ 2

\(\frac{ax-b}{a}+(a+b+1)x>\frac{2b}{a}\)

<=> \(x-\frac{b}{a}+\left(a+b+1\right)x>\frac{2b}{a}\)

<=> \(\left(a+b+2\right)x>\frac{3b}{a}\)

Giờ biện luận theo  a và b thôi

x=-1 với a=1.