đk a >= 0 ; a khác 9 

\(=\dfrac{a-3\sqrt{a}+2a+6\sqrt{a}-3a-9}{a-9}=\dfrac{3\sqrt{a}-9}{a-9}=\dfrac{3}{\sqrt{a}+3}\)

a) Để A xác định 

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-4\ne0\\\sqrt{x}-2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Vậy với \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\) thì A xác định 

b) \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(\Leftrightarrow A=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(\Leftrightarrow A=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Leftrightarrow A=\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Leftrightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để \(A=\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{-1}{3}\)

\(\Leftrightarrow3\sqrt{x}=-\sqrt{x}+2\)

\(\Leftrightarrow4\sqrt{x}=2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

+) \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)

\(=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)

\(=\sqrt{9^2-\left(\sqrt{17}\right)^2}\)

\(=8\)

+) \(3\sqrt{17}\approx12,4\)

\(\Rightarrow3\sqrt{17}>\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\)

e chịu

Ta có ^BAC = 180⁰ - ^B - ^C = 80⁰ 

Theo định lí sin \(\dfrac{BC}{sinA}=2R\Rightarrow2R\approx4,06\)

\(\dfrac{AC}{sinB}=2R\Rightarrow AC=2R.sinB\approx3,52\)

\(S_{ABC}=\dfrac{1}{2}.AC.BC.sinC\approx4,52\)( đvdt ) 

Điều kiện : \(x\ge0\)

 \(P=\dfrac{x}{\sqrt{x}+3}\)

Ta có : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+3\ge3\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{\sqrt{x}+3}\ge0\) hay \(P\ge0\)

Dấu bằng xảy ra \(\Leftrightarrow x=0\)

A = \(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}\) - \(\sqrt{5}\)

A =  -\(\sqrt{2}\)(\(\dfrac{1-\sqrt{3}}{1-\sqrt{3}}\)) - \(\sqrt{5}\)

A = -\(\sqrt{2}\) - \(\sqrt{5}\)