a)\(x\in R\)

b)\(x\ne1\)

c) \(x\notin\left\{1;2\right\}\)

d) \(x\notin\left\{3;-3\right\}\)

e) \(x\ne1\)

f) \(x\notin\left\{2;3\right\}\)

a) 

b) 

c) 

d) 

e) 

f) 

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

c: \(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}=\dfrac{-2}{x^2}\)

1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)

  Vậy ...

2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)

  Vậy ...

3) PT \(\Rightarrow3x-4-2x+5=3\)

          \(\Rightarrow x=2\)

 Vậy ...

4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

  Vậy ...

3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)

\(\Leftrightarrow3x-4-2x+5=3\)

\(\Leftrightarrow x+1=3\)

hay x=2

a: ĐKXĐ: x^2-2x<>0 và x^2-1>0

=>(x>1 và x<>2) hoặc x<-1

b: ĐKXĐ: x+1>0 và 5-3x>0

=>x>-1 và 3x<5

=>-1<x<5/3

c: DKXĐ: 5x+3>=0 và 3-x>0

=>x>=-3/5 và x<3

=>-3/5<=x<3

d: ĐKXĐ: 4-x^2>0 và 1+x>=0

=>x^2<4 và x>=-1

=>-2<x<2 và x>=-1

=>-1<=x<2

e: ĐKXĐ: 2-3x<>0 và 1-6x>0

=>x<>2/3 và x<1/6

=>x<1/6

\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)

a: \(P=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{x-1}\)

\(=\dfrac{1-1-3x}{3x}\cdot\dfrac{2x}{x-1}\)

\(=\dfrac{-3x}{3x}\cdot\dfrac{2x}{x-1}=\dfrac{-2x}{x-1}\)

Không có dấu "=" hay như nào đâu giải tìm x được

\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)