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阅读材料:像\((\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})=3\)、\(\sqrt{a}\cdot \sqrt{a}=a\) \((a\geqslant 0)\)、\((\sqrt{b}+1)(\sqrt{b}-1)=b-1\ \ (b\geqslant 0)……\)两个含有二次根式的代数式相乘,积不含有二次根式,我们称这两个代数式互为有理化因式\(.\)例如,\(\sqrt{3}\)与\(\sqrt{3}\)、\(\sqrt{2}+1\)与\(\sqrt{2}-1\)、\(2\sqrt{3}+3\sqrt{5}\)与\(2\sqrt{3}-3\sqrt{5}\)等都是互为有理化因式\(.\) 在进行二次根式计算时,利用有理化因式,可以化去分母中的根号.
例如;\(\dfrac{1}{2\sqrt{3}}=\dfrac{\sqrt{3}}{2\sqrt{3}\times \sqrt{3}}=\dfrac{\sqrt{3}}{6}\);\(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=\dfrac{{{(\sqrt{2}+1)}^{2}}}{(\sqrt{2}-1)(\sqrt{2}+1)}=3+2\sqrt{2}\).
解答下列问题:
\((1)\).\(3-\sqrt{7}\)与________互为有理化因式,将\(\dfrac{2}{3\sqrt{2}}\)分母有理化得________;
\((2)\)计算:\(\dfrac{1}{2-\sqrt{3}}-\dfrac{6}{\sqrt{3}}\);
\((3)\)己知有理数\(a\)、\(b\)满足\(\dfrac{a}{\sqrt{2}+1}+\dfrac{b}{\sqrt{2}}=-1+2\sqrt{2}\),求\(a\)、\(b\)的值.