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}

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)

Bài 5.5:

\(\left(2x-3\right)\left(x+1\right)+\left(4x^3-6x^2-6x\right):\left(-2x\right)=18\)

\(\Leftrightarrow\left(2x^2+2x-3x-3\right)+2x\cdot\left(2x^2-3x-3\right):\left(-2x\right)=18\)

\(\Leftrightarrow2x^2-x-3-2x^2+3x+3=18\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=\dfrac{18}{2}\)

\(\Leftrightarrow x=9\) 

Chọn đáp án C.

1: \(=\dfrac{2x^4-2x^2-3x^3-3x+6x^2-6+7}{x^2-1}\)

\(=2x^2-3x+6+\dfrac{7}{x^2-1}\)