a: \(\sqrt{\left(4-3\sqrt2\right)^2}=\left|4-3\sqrt2\right|=3\sqrt2-4\)
b: \(\sqrt{\left(2+\sqrt5\right)^2}=\left|2+\sqrt5\right|=\sqrt5+2\)
c: \(\sqrt{\left(4+\sqrt2\right)^2}=\left|4+\sqrt2\right|=4+\sqrt2\)
d: \(\sqrt{6-2\sqrt5}=\sqrt{\left(\sqrt5-1\right)^2}=\left|\sqrt5-1\right|=\sqrt5-1\)
e: \(\sqrt{7+4\sqrt3}\)
\(=\sqrt{\left(2+\sqrt3\right)^2}=\left|2+\sqrt3\right|=2+\sqrt3\)
f: \(\sqrt{12-6\sqrt3}\)
\(=\sqrt{\left(3-\sqrt3\right)^2}=\left|3-\sqrt3\right|=3-\sqrt3\)
h: \(\sqrt{17+12\sqrt2}\)
\(=\sqrt{9+2\cdot3\cdot2\sqrt2+8}\)
\(=\sqrt{\left(3+2\sqrt2\right)^2}=3+2\sqrt2\)
g: \(\frac{\sqrt2-\sqrt{11+6\sqrt2}}{\sqrt{6+2\sqrt5}-\sqrt5}\)
\(=\frac{\sqrt2-\sqrt{\left(3+\sqrt2\right)^2}}{\sqrt{\left(\sqrt5+1\right)^2}-\sqrt5}=\frac{\sqrt2-3-\sqrt2}{\sqrt5+1-\sqrt5}\)
=-3/1
=-3
i: \(\sqrt{6+2\cdot\sqrt{4-2\sqrt3}}\)
\(=\sqrt{6+2\cdot\left(\sqrt3-1\right)}\)
\(=\sqrt{4+2\sqrt3}\)
\(=\sqrt{\left(\sqrt3+1\right)^2}=\sqrt3+1\)
