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\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)
=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)
=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)
=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)
\(=(3-\sqrt{2}).\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2=7\)
\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)
\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)
\(=\sqrt{3-\sqrt{3}}\)
\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)
\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)
\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)
\(=2.6-18\sqrt{2}+16\sqrt{3}\)
\(=12-18\sqrt{2}+16\sqrt{3}\)
a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)
b)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)
a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)