\(A=\frac{x-2}{3x+2}\)

+A =0 => x -2 =0 => x =2

+ A<0  => (x-2)(3x+2) <0

=> x < -2/3  hoặc  x > 2

Ta có :

\(A=\frac{x-2}{3x+2}\)

a) \(A=0\Leftrightarrow\frac{x-2}{3x+2}=0\)

\(\Leftrightarrow x-2=0.\left(3x+2\right)\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

b) \(A< 0\Rightarrow\frac{x-2}{3x+2}< 0\)

\(\Rightarrow\)\(x-2< 0\) ; \(3x+2>0\)   hoặc   \(x-2>0\) ; \(3x+2< 0\)

\(\Rightarrow\)\(x< 2\); \(3x>-2\)  hoặc   \(x>2\) ; \(3x< -2\)

\(\Rightarrow\)\(x< 2\); \(x>\frac{-2}{3}\) hoặc  \(x>2\) ; \(x< \frac{-2}{3}\)

\(\Rightarrow\frac{-2}{3}< x< 2\) hoặc \(x\in\varnothing\)

Vậy \(-\frac{2}{3}< x< 2\) thì \(A< 0\)

a)   Xét A = 0

\(\Leftrightarrow\frac{X-2}{3X+2}=0\)

\(\Leftrightarrow X-2=0\)

\(\Leftrightarrow X=2\)

b)  Xét A < 0

\(\Leftrightarrow\frac{X-2}{3X+2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}X-2< 0\\3X+2< 0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X< 1\\X< -1\end{cases}}\)

Trả lời:

a,  \(ĐK:x\ne\frac{1}{3}\)

 \(A=\frac{3x+1-1}{1-3x}:\frac{3x-9x^2}{3x-1}=\frac{3x}{1-3x}\cdot\frac{3x-1}{3x-9x^2}=\frac{3x.\left(3x-1\right)}{\left(1-3x\right)\left(3x-9x^2\right)}=\frac{3x\left(3x-1\right)}{\left(1-3x\right)3x\left(1-3x\right)}\)

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

b, \(5x^2+3x=0\)

\(\Leftrightarrow x\left(5x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}}\)

Thay x = 0 vào A, ta có :

\(A=\frac{1}{3.0-1}=\frac{1}{-1}=-1\)

Thay x = - 3/5 vào A, ta có :

\(A=\frac{1}{3.\left(-\frac{3}{5}\right)-1}=\frac{1}{-\frac{9}{5}-1}=\frac{1}{-\frac{14}{5}}=-\frac{5}{14}\)

c, \(A=\frac{x}{x-1}\)

\(\Leftrightarrow\frac{1}{3x-1}=\frac{x}{x-1}\)\(\left(ĐK:x\ne\frac{1}{3};x\ne1\right)\)

\(\Leftrightarrow\frac{x-1}{\left(3x-1\right)\left(x-1\right)}=\frac{x\left(3x-1\right)}{\left(3x-1\right)\left(x-1\right)}\)

\(\Rightarrow x-1=3x^2-x\)

\(\Leftrightarrow3x^2-x-x+1=0\)

\(\Leftrightarrow3x^2-2x+1=0\)

\(\Leftrightarrow3\left(x^2-\frac{2}{3}x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow x^2-\frac{2}{3}x+\frac{1}{3}=0\)

\(\Leftrightarrow x^2-2.x.\frac{1}{3}+\frac{1}{9}+\frac{2}{9}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2+\frac{2}{9}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=-\frac{2}{9}\) (vô lí)

Vậy không tìm được x thỏa mãn đề bài.

d, \(\frac{6}{A}=\frac{6}{\frac{1}{3x-1}}=6\left(3x-1\right)=18x-6\)

Vậy x thuộc Z thì 6/A thuộc Z

\(A=\left(3x+1-\frac{1}{1-3x}\right):\left(\frac{3x-9x^2}{3x-1}\right)=\left(\frac{1-9x^2-1}{1-3x}\right):\left(\frac{3x\left(1-3x\right)}{3x-1}\right)=-\frac{9x}{1-3x}:\left(-3x\right)=\frac{3}{1-3x}\)

b. Với \(5x^2+3x=0\Leftrightarrow x\left(5x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\) nhưng mà ở trên ta cần có điều kiện x#0 nên

\(x=-\frac{3}{5}\Rightarrow A=\frac{3}{1-3\times\left(-\frac{3}{5}\right)}=\frac{15}{14}\)

c.\(A=\frac{x}{x-1}=\frac{3}{1-3x}\Leftrightarrow x-3x^2=3x-3\Leftrightarrow3x^2+2x-3=0\Leftrightarrow x=\frac{-1\pm\sqrt{10}}{3}\)

d.\(\frac{6}{A}=2\times\left(1-3x\right)\) nguyên nên \(1-3x=-\frac{k}{2}\Leftrightarrow x=\frac{k+2}{6}\) với k là số nguyên 

a)

\(A=\frac{x^2-2x-x+2}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+2}\)

b)

+A> 0  => x>1 hoặc x < -2

+ A<0 => -2 <x<1

+A =0 => x =1

+A có nghĩa khi  x khác 2 và -2

+A vô nghĩa khi x =2; x =-2