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 \(x\in\left\{3;1;5;-1\right\}\)

a: Khi x=16 thì \(A=\dfrac{2\cdot\sqrt{16}}{\sqrt{16}+3}=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)

b: P=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{7\sqrt{x}+3}{9-x}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+5\sqrt{x}+6}{x-9}\)

a: Thay x=16 vào A, ta được:

\(A=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)

Ta có: M=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

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

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

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

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

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

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

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

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

a)

\(2x\left(9-x\right)+\left(2x+5\right)\left(x+1\right)\\ =18x-2x^2+2x^2+5x+2x+5\\ =25x+5\\ =5\left(5x+1\right)\)

b)

\(\left(x-3\right)^2+\left(x+3\right)^2+2\left(3-x\right)\left(3+x\right)\\ =\left(3-x\right)^2+2\left(3-x\right)\left(3+x\right)+\left(3+x\right)^2\\ =\left[\left(3-x\right)+\left(3+x\right)\right]^2\\ =6^2=36\)

`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`

Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`

`b)` Với `x >= 0,x ne 4` có:

`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[\sqrt{x}-2]/[\sqrt{x}-3]`

`c)` Với `x >= 0,x ne 4` có:

`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`

Có: `C >= 1`

`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`

`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`

`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`

  Vì `x >= 0=>\sqrt{x} >= 0`

  `=>\sqrt{x}-3 > 0`

`<=>x > 9` (t/m đk)