a) 

Với A=0

\(\Rightarrow x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

với A<0

\(\Rightarrow x\left(x-4\right)< 0\)

\(th1\orbr{\begin{cases}x< 0\\x-4>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 0\\x>4\end{cases}\Leftrightarrow4< x< 0\left(vl\right)}\)

\(th2\orbr{\begin{cases}x>0\\x-4< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>0\\x< 4\end{cases}\Leftrightarrow0< x< 4\left(tm\right)}\)

\(\Leftrightarrow0< x< 4\Leftrightarrow x\in\left\{1;2;3\right\}\)

Với A>0

\(\Rightarrow x\left(x-4\right)>0\)

\(th1\orbr{\begin{cases}x>0\\x-4>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>0\\x>4\end{cases}}\Leftrightarrow x>4\)

\(th2\orbr{\begin{cases}x< 0\\x-4< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 0\\x< 4\end{cases}}\Leftrightarrow x< 0\)

b) 

Với B=0

\(\Rightarrow\frac{x-3}{x}=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\Rightarrow x=3\\x=0\left(l\right)\end{cases}}\)

vậy x=3 thì B = 0

Với B < 0

\(\Rightarrow\frac{x-3}{x}< 0\)

\(th1\orbr{\begin{cases}x-3>0\\x< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>3\\x< 0\end{cases}\Leftrightarrow3< x< 0\left(vl\right)}\)

\(th2\orbr{\begin{cases}x-3< 0\\x>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 3\\x>0\end{cases}\Leftrightarrow0< x< 3\left(tm\right)\Leftrightarrow x\in\left\{1;2\right\}}\)

Với B > 0

\(th1\orbr{\begin{cases}x-3>0\\x>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>3\\x>0\end{cases}\Leftrightarrow x>3}\)

\(th2\orbr{\begin{cases}x-3< 0\\x< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 0\end{cases}\Leftrightarrow x< 0}\)

bài1   A=\(\left(\frac{3-x}{x+3}\cdot\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

=\(\left(-\frac{x-3\cdot\left(x+3\right)^2}{\left(x+3\right)^2\cdot\left(x-3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

=\(-\frac{x}{x+3}\cdot\frac{x+3}{3x^2}=\frac{-1}{3x}\)

b)  thế \(x=-\frac{1}{2}\)vào biểu thức A

 \(-\frac{1}{3\cdot\left(-\frac{1}{2}\right)}=\frac{2}{3}\)

c)  A=\(-\frac{1}{3x}< 0\)

VÌ (-1) <0  nên  3x>0

                        x >0

1) a) Để x > 0

=> \(2a-5< 0\)

\(\Rightarrow2a< 5\)

\(\Rightarrow a< 2,5\)

\(\text{Vậy }x>0\Leftrightarrow a< 2,5\)

b) Để x < 0

\(\Rightarrow2a-5>0\)

\(\Rightarrow2a>5\)

\(\Rightarrow a>2,5\)

\(\text{Vậy }x< 0\Leftrightarrow a>2,5\)

c) Để x = 0

\(\Rightarrow2a-5=0\)

\(\Rightarrow2a=5\)

\(\Rightarrow a=2,5\)

\(\text{Vậy }x=0\Leftrightarrow a=2,5\)

2) \(\text{Vì }a\inℤ\Rightarrow3a-5\inℤ\)

\(\text{mà }x\inℤ\Leftrightarrow3a-5⋮4\)

\(\Rightarrow3a-5\in B\left(4\right)\)

\(\Rightarrow3a-5\in\left\{0;4;8;...\right\}\)

\(\Rightarrow3a\in\left\{5;9;13;....\right\}\)

\(\Rightarrow a\in\left\{\frac{5}{3};3;\frac{13}{3};6;....\right\}\)

\(\text{Mà }a\inℤ\Rightarrow a\in\left\{3;6;9;...\right\}\text{thì }x\inℤ\)