\(\sqrt{16-8\sqrt{3}}-\sqrt{16+8\sqrt{3}}=\sqrt{12-8\sqrt{3}+4}-\sqrt{12+8\sqrt{3}+4}\)

\(=\sqrt{\left(2\sqrt{3}-2\right)^2-\sqrt{\left(2\sqrt{3}+2\right)^2}}=2\sqrt{3}-2-2\sqrt{3}-2=-4\)

a) \(\sqrt{x+2\sqrt{x-1}}=2\)(ĐK: \(x\ge1\))

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=1\)

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

\(\Leftrightarrow\sqrt{x-1}+1=1\)

\(\Leftrightarrow x=1\)(thỏa mãn) 

b) \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

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

\(\Leftrightarrow\left|1-x\right|+\left|x+2\right|=3\)

Ta có: \(\left|1-x\right|+\left|x+2\right|\ge\left|1-x+x+2\right|=3\)\

Dấu \(=\)khi \(\left(1-x\right)\left(x+2\right)\ge0\Leftrightarrow-2\le x\le1\).

Vậy nghiệm phương trình đã cho là \(x\in\left[-2,1\right]\)

Đk: \(x\ge0\)

\(pt\Leftrightarrow7+\sqrt{2x}=\left(3+\sqrt{5}\right)^2\)

\(\Leftrightarrow7+\sqrt{2x}=8+6\sqrt{5}\)

\(\Leftrightarrow\sqrt{2x}=1+6\sqrt{5}\)

\(\Leftrightarrow2x=\left(1+6\sqrt{5}\right)^2\)

\(\Leftrightarrow2x=181+12\sqrt{5}\)

\(\Leftrightarrow x=90,5+6\sqrt{5}\)(tm) 

Xét tam giác \(ABC\)có \(AB\le AC\le BC\), phân giác \(AD\)ứng với cạnh lớn nhất, đường cao \(CH\)ứng với cạnh nhỏ nhất. 

Kẻ \(DK\perp AB\).

\(\frac{AB}{BD}=\frac{AC}{DC}\Rightarrow BD\le DC\).

Suy ra \(BD\le\frac{1}{2}BC\Rightarrow DK\le\frac{1}{2}CH\).

Vì tam giác \(ABC\)có \(AB\le AC\le BC\)nên \(\widehat{BAC}\ge60^o\Leftrightarrow\widehat{BAD}\ge30^o\).

Suy ra tam giác \(BKD\)có \(DK\ge\frac{1}{2}AD\).

Suy ra \(CH\ge AD\).

Ta có đpcm.