Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
trinh mai
\(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{3^2+3.2\sqrt{2}+2}=\sqrt{\left(3-\sqrt{2}\right)^2}.\sqrt{\left(3+\sqrt{2}\right)^2}=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)=3^2-2=7\)
\(a,\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+4\sqrt{5}+4}}=\sqrt{17-4\sqrt{\left(\sqrt{5}\right)^2+2.2\sqrt{5}+2^2}}=\sqrt{17-4\sqrt{\sqrt{\left(\sqrt{5}+2\right)^2}}}=\sqrt{17-4\sqrt{\sqrt{5}+2}}\) \(b,\sqrt{a};đk:a\ge0;2-3=-1< 0\Rightarrow sai\)
\(c,\sqrt{\left(\sqrt{3-3}\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}=\sqrt{0^2}.\sqrt{\frac{1}{3-\sqrt{3}}}=0.\sqrt{\frac{1}{3-\sqrt{3}}}=0\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}=\left(\sqrt{2}.\sqrt{3}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right)2\sqrt{6}=\left[\sqrt{3}\left(\sqrt{2}-3\right)+\sqrt{2}.4\right]2\sqrt{6}=\left[2.\sqrt{3}.\sqrt{2}.\sqrt{3}\left(\sqrt{2}-3\right)+\sqrt{2}.\sqrt{2}.\sqrt{3}.2.4\right]=6\sqrt{2}\left(\sqrt{2}-3\right)+16\sqrt{3}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)
\(=(3-\sqrt{2}).\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2=7\)
\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)
\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)
\(=\sqrt{3-\sqrt{3}}\)
\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)
\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)
\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)
\(=2.6-18\sqrt{2}+16\sqrt{3}\)
\(=12-18\sqrt{2}+16\sqrt{3}\)