a: 

Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)

\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)

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

b: x^2-4x+3=0

=>x=1(nhận) hoặc x=3(loại)

Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)

c: P>0

=>x-2>0

=>x>2

d: P nguyên

=>4x^2 chia hết cho x-2

=>4x^2-16+16 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}

=>x thuộc {1;4;6;-2;10;-6;18;-14}

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

a: \(P=\dfrac{x^2+6x+9-x^2+6x-9-4}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-1}{x-3}\)

\(=\dfrac{4\left(3x-1\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{3x-1}=\dfrac{4}{x+3}\)

a: ĐKXĐ: \(\left[{}\begin{matrix}x\le-\dfrac{3}{2}\\x>3\end{matrix}\right.\)

b: ĐKXĐ: x>3

c: Ta có: A=B

\(\Leftrightarrow\sqrt{\dfrac{2x+3}{x-3}}=\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)

\(\Leftrightarrow0x=0\)(luôn đúng với mọi x>3)

a: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

Ta có: \(A=\dfrac{x^3-3}{x^2-2x-3}+\dfrac{6-2x}{x+1}+\dfrac{x+3}{3-x}\)

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

\(=\dfrac{x^3-3-2x^2+12x-18-x^2-4x-3}{\left(x-3\right)\left(x+1\right)}\)

\(=\dfrac{x^4-3x^2+8x-24}{\left(x-3\right)\left(x+1\right)}\)

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

\(=\dfrac{x^2+8}{x+1}\)

b: Ta có: A=x-2

\(\Leftrightarrow x^2+8=x^2-x-2\)

\(\Leftrightarrow8+x+2=0\)

hay x=-10

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)

\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{1}{4}\)

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{3}-\frac{1}{4}\)

\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)

\(\frac{2}{3}-x=\frac{1}{12}-\frac{5}{4}\)

\(\frac{2}{3}-x=-\frac{7}{6}\)

\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)

\(x=\frac{2}{3}+\frac{7}{6}\)

\(x=\frac{11}{6}\)

Thay x=28 và A=x+|10|

ta có 

A=28+|10|

A=28+10

A=38