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Ta có: \(A=\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\frac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}\right)^2-\left(\sqrt{5-3\sqrt{2}}\right)^2}-\frac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{\left(\sqrt{3+\sqrt{2}}\right)^2-\left(\sqrt{3-\sqrt{2}}\right)^2}\)
\(=\frac{3\left[\left(\sqrt{5+3\sqrt{2}}\right)^2+2\cdot\sqrt{5+3\sqrt{2}}\cdot\sqrt{5-3\sqrt{2}}+\left(\sqrt{5-3\sqrt{2}}\right)^2\right]}{\left|5+3\sqrt{2}\right|-\left|5-3\sqrt{2}\right|}-\frac{\left(\sqrt{3+\sqrt{2}}\right)^2+2\cdot\sqrt{3+\sqrt{2}}\cdot\sqrt{3-\sqrt{2}}+\left(\sqrt{3-\sqrt{2}}\right)^2}{\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|}\)
\(=\frac{3\left(\left|5+3\sqrt{2}\right|+2\sqrt{7}+\left|5-3\sqrt{2}\right|\right)}{5+3\sqrt{2}-\left(5-3\sqrt{2}\right)}-\frac{\left|3+\sqrt{2}\right|+2\cdot\sqrt{7}+\left|3-\sqrt{2}\right|}{3+\sqrt{2}-\left(3-\sqrt{2}\right)}\)
\(=\frac{3\left(5+3\sqrt{2}+2\sqrt{7}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\frac{3+\sqrt{2}+2\sqrt{7}+3-\sqrt{2}}{3+\sqrt{2}-3+\sqrt{2}}\)
\(=\frac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\frac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(=\frac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\frac{3\left(6+2\sqrt{7}\right)}{6\sqrt{2}}\)
\(=\frac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}\)
\(=\frac{12}{6\sqrt{2}}=\sqrt{2}\)
Vậy: \(A=\sqrt{2}\)
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)
Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3
\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)
\(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
mik chỉnh lại đề
\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)
