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\(A=\sqrt{2+2\sqrt{\frac{3}{4}}}+\sqrt{2-2\sqrt{\frac{3}{4}}}\)
\(A=\sqrt{\left(\sqrt{\frac{3}{2}}\right)^2+2\sqrt{\frac{3}{2}.\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{3}{2}}\right)^2-2\sqrt{\frac{3}{2}.\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)
\(A=\sqrt{\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)^2}\)
\(A=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}-\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\)
\(A=2\sqrt{\frac{3}{2}}=\sqrt{4.\frac{3}{2}}=\sqrt{6}\)
\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=|2+\sqrt{3}|+|2-\sqrt{3}|\)\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(\frac{4}{\sqrt{3}+1}-\frac{5}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}\)
\(=\frac{4\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\frac{5\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{6\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(\sqrt{3}-3\right)}\)
\(=\frac{4\sqrt{3}-4}{2}-\frac{5\sqrt{3}+10}{-1}+\frac{6\sqrt{3}+18}{3-9}\)
\(=2\sqrt{3}-2+5\sqrt{3}+10-\sqrt{3}-3\)
\(=6\sqrt{3}+5\)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)
b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)