\(x^2\left(1-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

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

\(=x^2+27\)

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 \(x\in\left\{3;1;5;-1\right\}\)

\(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

\(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)

\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{6}{3-x}.\dfrac{x\left(x-3\right)}{x+1}\) \(=\dfrac{6x}{x+1}\)

9(x + 1)² - 16(y + 3)²

= 9(x² + 2x + 1) - 16(y² + 6y + 9)

= 9x² + 18x + 9 - 16y² - 96y - 144

= 9x² - 16y² + 18x - 96y - 135

\(C=\dfrac{21+\left(x^2-x-12\right)-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+3-1}{x+3}\)

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

\(A=\left(\dfrac{3x-x^2}{9-x^2}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}+\dfrac{x-3}{2-x}-\dfrac{x+2}{x+3}\right)\left(dk:x\ne\pm3,x\ne2\right)\)

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

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

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

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

\(=\dfrac{-3\left(x-2\right)}{22-3x^2}\)

\(=\dfrac{-3x+6}{22-3x^2}\)

Vậy \(A=\dfrac{-3x+6}{22-3x^2}\) với \(x\ne\pm3,x\ne2\)

Ta có: \(\left(\dfrac{x^2-3x}{x^2-9}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\right)\)

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

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

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

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

Điều kiện : x ≠ 2 ; x ≠ 3 ; x ≠ - 3

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

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

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

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

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