5.
观察、思考、解答:
\(( \sqrt {2}-1)^{2}=( \sqrt {2})^{2}-2×1× \sqrt {2}+1^{2}=2-2 \sqrt {2}+1=3-2 \sqrt {2}\)
反之\(3-2 \sqrt {2}=2-2 \sqrt {2}+1=( \sqrt {2}-1)^{2}\)
\(∴3-2 \sqrt {2}=( \sqrt {2}-1)^{2}\)
\(∴ \sqrt {3-2 \sqrt {2}}= \sqrt {2}-1\)
\((1)\)仿上例,化简:\( \sqrt {6-2 \sqrt {5}}\);
\((2)\)若\( \sqrt {a+2 \sqrt {b}}= \sqrt {m}+ \sqrt {n}\),则\(m\)、\(n\)与\(a\)、\(b\)的关系是什么?并说明理由;
\((3)\)已知\(x= \sqrt {4- \sqrt {12}}\),求\(( \dfrac {1}{x-2}+ \dfrac {1}{x+2})⋅ \dfrac {x^{2}-4}{2(x-1)}\)的值\((\)结果保留根号\()\)