(x + 2)(x – 2) – (x – 3)(x + 1)

= x2 – 22 – (x2 + x – 3x – 3)

= x2 – 4 – x2 – x + 3x + 3

= 2x – 1

chịu

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

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

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

           \(=\frac{1}{x\left(x+1\right)}\)

Chúc bạn học tốt !!!

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

=> A = \(\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{x+1}{x\left(x-1\right)}\)

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

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

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

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

=> A = \(\frac{1}{x\left(x+1\right)}\) (Đk: x \(\ne\)0 hoặc x \(\ne\)-1)

\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)

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

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

à xin lỗi mình nhầm dòng cuối 

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

Để biểu thức trên nhận giá trị dương khi 

\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi 

`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`

\(=2x^2-2-2x^2+8x+3x-12\)

=11x-14

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 \(x\in\left\{3;1;5;-1\right\}\)