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

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

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

\(=8\)

+) \(3\sqrt{17}\approx12,4\)

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

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 = -\(\sqrt{2}\) - \(\sqrt{5}\)

Mỗi bạn có 10 viên bi hồi tring em nhé chúc em học tốt 

đk x >= 0 ; x khác 4 ; 9 

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

b, \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

c, \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-2\ge0\Leftrightarrow\dfrac{\sqrt{x}+1-2\sqrt{x}+4}{\sqrt{x}-2}\ge0\)

\(\Leftrightarrow\dfrac{5-\sqrt{x}}{\sqrt{x}-2}\ge0\Leftrightarrow\left\{{}\begin{matrix}x\le25\\x>4\end{matrix}\right.\Leftrightarrow4< x\le25\)

Kết hợp đk vậy 4 < x =< 25  ; x khác 9