a: Thay x=-4 vào B, ta được:

\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)

b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)

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

c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)

`Answer:`

`a)`

`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`

`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`

`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`

`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`

`=>A=-2x^2+28x-6`

`b)`

`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`

`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`

`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`

`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`

Thay `x=-7` vào ta được:

`B=10(-7)^2-2(-7)^3-7(-7)-6`

`=>B=10.49-2(-343)+49-6`

`=>B=490+686+49-6`

`=>B=1219`

a: ĐKXĐ: x∉{3;-3;2}

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

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

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

b: \(x^3-4x=0\)

=>x*(x-2)(x+2)=0

=>x∈{0;2;-2}

mà x∉{3;-3;2}

nên x=0 hoặc x=-2

Khi x=0 thì \(P=\frac{3}{x-2}=\frac{3}{0-2}=-\frac32\)

Khi x=-2 thì \(P=\frac{3}{x-2}=\frac{3}{-2-2}=-\frac34\)

dell hiểu đc đề ntn cả?

ngu nên đéo hiểu

a p=1/x-3

b tự làm

Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)