\(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}-3\right)^2}{2a-\sqrt{a}}\)
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\(\sqrt{x-1}\le2\\ < =>0\le x-1\le4\\ < =>1\le x\le5\)
\(\sqrt{3x-2}=2-x\left(x\ge\dfrac{2}{3}\right)\\ < =>\left\{{}\begin{matrix}2-x\ge0\\3x-2=\left(2-x\right)^2=x^2-4x+4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}-x\ge-2\\x^2-7x+6=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le2\\\left(x-6\right)\left(x-1\right)=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le2\\\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\end{matrix}\right.< =>x=1\left(TM\right)\)
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\(\sqrt{3+2\sqrt{2}}\) + \(\sqrt{6-4\sqrt{2}}\)
= \(\sqrt{2+2\sqrt{2}+1}\) + \(\sqrt{4-4\sqrt{2}+2}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}\) + \(\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}\) + 1 + 2 - \(\sqrt{2}\)
= 3
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đk a > 0
\(=\dfrac{4+4\sqrt{a}+a-a+6\sqrt{a}-9}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{10\sqrt{a}-5}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{5\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{5}{\sqrt{a}}\)