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Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(a,\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+\sqrt{1-2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}}\)
\(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{13+6\sqrt{1+2\sqrt{2}+2}}\)
\(=\sqrt{13+6\sqrt{\left(1+\sqrt{2}\right)^2}}\)
\(=\sqrt{13+6\left(1+\sqrt{2}\right)}=\sqrt{13+6+\sqrt{12}}\)
\(=\sqrt{19+2\sqrt{3}}\)
a) = \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
= \(\sqrt{13+6\sqrt{4+\sqrt{8-2.2\sqrt{2}+1}}}\)
= \(\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)
= \(\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
= \(\sqrt{13+6\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+6\left(\sqrt{2}+1\right)}\)
= \(\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}\)
= \(\sqrt{18+2.3\sqrt{2}+1}\)
= \(\sqrt{\left(3\sqrt{2}+1\right)^2}\)
= \(3\sqrt{2}+1\)