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\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(2-2\sqrt{2}.\sqrt{3}+3\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=3-2=1\)
\(A=\dfrac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\\ =\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\\ =\dfrac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{3}-\sqrt{2}}\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ =3-2\\ =1\)
Vậy \(A=1\)
\(VT=\frac{\left(5\sqrt{3}+5\sqrt{2}\right).\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)^2.\left(5-2\sqrt{6}\right)}{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5\sqrt{3}-5\sqrt{2}\right)}\)\(=\frac{\left(75+50\sqrt{6}+50\right).\left(5-2\sqrt{6}\right)}{75-50}\)
\(=\frac{25\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)}{25}=5^2-\left(2\sqrt{6}\right)^2\)\(=25-24=1=VP\)
bn chép lại đề nhé
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{\left(75+50\sqrt{6}+50\right)\left(\sqrt{3}-\sqrt{2}\right)}{75-50}\)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\frac{\left(5\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(3-2.\sqrt{3}.\sqrt{2}+2\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1\)
DÀI QUÁ MK KO GHI ĐƯỢC NÊN VIẾT KQ LUÔN NHA !!!
ĐẲNG THỨC ĐÓ = 1 NHA 
Hatsune Miku !
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
\(a=\dfrac{4\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}}{2}\)
\(=2\sqrt{\sqrt{5}-\sqrt{5}+1}=2\)
\(P=\left(2^5-7\cdot2^2-3\right)^{81}+19=1+19=20\)
\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)
\(=(3-\sqrt{2}).\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2=7\)
\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)
\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)
\(=\sqrt{3-\sqrt{3}}\)
\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)
\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)
\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)
\(=2.6-18\sqrt{2}+16\sqrt{3}\)
\(=12-18\sqrt{2}+16\sqrt{3}\)
cm > hay < ?