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\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
A=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{5+2\sqrt{5}.\sqrt{3}+3}\)
A=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
A=\(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
A=\(-2\sqrt{3}\)
\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(A=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(A=\left|\sqrt{5}-\sqrt{3}\right|-\sqrt{5}-\sqrt{3}\)
\(A=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(A=-2\sqrt{3}\)
\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{1}{2}\) NHA 
Nguyễn Thị My Na !
\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{3-2\sqrt{3.5}+5}-\sqrt{3+2\sqrt{3.5}+5}\)
\(=\sqrt{\left(3-5\right)^2}-\sqrt{\left(3+5\right)^2}\)
\(=|3-5|-|3+5|\)
\(=-3+5-3-5\)
\(=-6 \)
a)
\(\left(3-\sqrt{15}\right)\sqrt{4+\sqrt{15}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{5+2\sqrt{15}+3}}{\sqrt{2}}\\ =\left(3-\sqrt{15}\right)\cdot\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}\\ =\left(\sqrt{9}-\sqrt{15}\right)\cdot\dfrac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}\)
\(=\sqrt{3}\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\) (vì \(\sqrt{5}+\sqrt{3}>0\))
\(=\sqrt{3}\cdot\dfrac{3-5}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-2}{\sqrt{2}}\\ =\sqrt{3}\cdot\dfrac{-\sqrt{4}}{\sqrt{2}}\\ =-\sqrt{6}\)
b)
\(\sqrt{29-12\sqrt{5}}-\sqrt{24-8\sqrt{5}}\\ =\sqrt{20-2\cdot3\cdot2\sqrt{5}+9}-\sqrt{20-2\cdot2\cdot2\sqrt{5}+4}\\ =\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(2\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}-3\right|-\left|2\sqrt{5}-2\right|\)
\(=2\sqrt{5}-3-\left(2\sqrt{5}-2\right)\) (vì \(2\sqrt{5}-3>0;2\sqrt{5}-2>0\))
\(=2\sqrt{5}-3-2\sqrt{5}+2\\ =-1\)
\(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{4}+\sqrt{2}\right)-\left(\sqrt{3}+\sqrt{6}\right)+\left(\sqrt{4}+\sqrt{8}\right)}{2+\sqrt{2}-\sqrt{3}}\) ( Tách 4 thành \(\sqrt{4}+\sqrt{4}\) )
\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)-\sqrt{3}\left(1+\sqrt{2}\right)+\sqrt{4}\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{3}+2\right)\left(\sqrt{2}+1\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\sqrt{2}+1\)
\(=\frac{2+\sqrt{2}-\sqrt{3}+2-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}=1+\frac{\sqrt{2}\left(2+\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}=1+\sqrt{2}\)
Ta có:
\(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{2+\sqrt{2}-\sqrt{3}+2-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
\(=1+\frac{\sqrt{2}(2+\sqrt{2}-\sqrt{3})}{2+\sqrt{2}-\sqrt{3}}\)
\(=1+\sqrt{2}\)
Vậy \(\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}=1+\sqrt{2}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\frac{1}{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\sqrt{3}-2-\sqrt{2}=-2\)
\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{2}\sqrt{8-\sqrt{15}}}{\sqrt{2}\left(\sqrt{15}.\sqrt{2}-\sqrt{2}\right)}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}.\sqrt{2}\left(\sqrt{15}-1\right)}\)
\(=\frac{\sqrt{15-2\sqrt{15}+1}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)