Ta có điều kiện xác định \(x\ge0\)

xét \(x-3\sqrt{x}+5=x-\frac{2.3}{2}\sqrt{x}+\frac{9}{4}+\frac{11}{4}\)

\(=\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Do đó \(\frac{1}{x-3\sqrt{x}+5}\le\frac{4}{11}\).Vậy GTNN của biểu thức là \(\frac{4}{11}\)khi \(\sqrt{x}-\frac{3}{2}=0\)hay \(x=\frac{9}{4}\)

Biểu thức trên không có GTLN

\(\frac{1}{\sqrt{3}-1}-\frac{1}{2}=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{1}{2}\)

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

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

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

\(=\frac{1\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{1}{2}\)      

\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}\right)^2-1^2}-\frac{1}{2}\)   

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

\(=\frac{\sqrt{3}}{2}\)

\(\sqrt{11+4\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}-\sqrt{2-2\sqrt{2}.\sqrt{3}+3}\)

\(=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

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

\(=3\sqrt{2}\)

\(\sqrt{11+4\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{8+4\sqrt{6}+3}-\sqrt{3-2\sqrt{6}+2}\)

\(=\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\left|2\sqrt{2}+\sqrt{3}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)

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

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

\(=3\sqrt{2}\)

\(3\sin^2\alpha+2\sin\alpha=1\Leftrightarrow3\sin^2\alpha+3\sin\alpha-\sin\alpha-1=0\Leftrightarrow3\sin\alpha\left(\sin\alpha+1\right)-\left(\sin\alpha+1\right)=0\Leftrightarrow\left(3\sin\alpha-1\right)\left(\sin\alpha+1\right)=0\)Do \(\sin\alpha>0\)nên \(\sin\alpha+1>0\)suy ra \(3\sin\alpha-1=0\Leftrightarrow\sin\alpha=\frac{1}{3}\)

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}=\frac{2\sqrt{2}}{3}\)

\(\Rightarrow\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=\frac{\sqrt{2}}{4}\)