a: 

Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)

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

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

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

b: P>=1/2

=>P-1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)

=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)

=>\(-\sqrt{x}-15>=0\)

=>\(-\sqrt{x}>=15\)

=>căn x<=-15

=>\(x\in\varnothing\)

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>P>=-2

Dấu = xảy ra khi x=0

\(P=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{x-9}:\dfrac{1}{\sqrt{x}-3}\)

    \(=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}-3\right)=\dfrac{6}{\sqrt{x}+3}\)

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

\(P=\dfrac{6}{\sqrt{x}+3}\)

Bài 1: 

a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)

Do đó: A>=0

`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

ĐKXĐ: \(x\ge0;x\ne9\)

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

\(P=\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right)\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)

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

\(P=\dfrac{-3}{\sqrt{x}+3}\)

b/ Do \(-3< 0\Rightarrow P_{min}\) khi \(\sqrt{x}+3\) nhỏ nhất

Mà \(\sqrt{x}+3\ge3\Rightarrow P_{min}=\dfrac{-3}{3}=-1\) khi \(\sqrt{x}+3=3\Leftrightarrow x=0\)

Vậy với \(x=0\) thì P đạt GTNN

a) \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)

b) Ta có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\)

Dấu bằng xảy ra khi x=0

Vậy x=0 thì P đạt GTNN là -1

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

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

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