a) ĐKXĐ: \(a\ge0;a\ne1\)

\(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)

\(=\left[\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}\right]\)

\(=\left(1+2\sqrt{a}+a\right)\left(1-2\sqrt{a}+a\right)\)

\(=\left(1-a\right)^2\)

b) Để \(P< 7-4\sqrt{3}\)

\(\Rightarrow\left(1-a\right)^2< 7-4\sqrt{3}\)

\(\Leftrightarrow\left|1-a\right|< \left(2-\sqrt{3}\right)^2\)

\(\Leftrightarrow\sqrt{3}-2< a-1< 2-\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}-1< a< 3-\sqrt{3}\)

Vậy \(\sqrt{3}-1< a< 3-\sqrt{3}\) thì \(P< 7-4\sqrt{3}\)

`a)P=((1-asqrta)/(1-sqrta)+sqrta).((1+asqrta)/(1+sqrta)-sqrta)`

`=(((1-sqrta)(a+sqrta+1))/(1-sqrta)+sqrta).(((1+sqrta)(a-sqrta+1))/(1+sqrta)-sqrta)`

`=(a+sqrta+1+sqrta)(a-sqrta+1-sqrta)`

`=(a+2sqrta+1)(a-2sqrta+1)`

`=(sqrta+1)^2(sqrta-1)^2`

`=(a-1)^2`

`b)a<7-4sqrt3`

`<=>(a-1)^2<(2-sqrt3)^2`

`<=>sqrt3-2<a-1<2-sqrt3`

`<=>sqrt3-1<a<3-sqrt3`

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

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

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

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

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

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

c) Để \(P< -\dfrac{1}{2}\) thì \(P+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)

a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)

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

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

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

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

\(=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\) khi 

\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)

\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)

\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)