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1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)
\(=5\sqrt{3}\)
2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)
\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)
\(=\dfrac{-\sqrt{3}-5}{11}\)
3) Ta có: \(\sqrt{\dfrac{2}{5}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)
\(=\dfrac{\sqrt{10}}{5}\)
Nếu em thấy các câu hỏi do lag mà bị gửi đúp (tức là rất nhiều câu hỏi giống nhau xuất hiện cùng 1 chỗ) thì xóa giúp mình nhé cho đỡ vướng. Nhưng nhớ để lại 1 câu. Cảm ơn em.
\(\dfrac{4}{\sqrt{5}-\sqrt{2}}+\dfrac{3}{\sqrt{5}-2}-\dfrac{2}{\sqrt{3}-2}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{5}\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^2-2^2}-\dfrac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^2-2^2}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\left(\sqrt{2}+\sqrt{5}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{8\left(\sqrt{2}+\sqrt{5}\right)}{6}+\dfrac{18\left(\sqrt{5}+2\right)}{6}+\dfrac{12\left(\sqrt{3}+2\right)}{6}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{8\sqrt{2}+8\sqrt{5}+18\sqrt{5}+36+12\sqrt{3}+24-\sqrt{3}+1}{6}\)
\(=\dfrac{8\sqrt{2}+26\sqrt{5}+11\sqrt{3}+61}{6}\)
\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}+\dfrac{2\left(2+\sqrt{3}\right)}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{4\sqrt{5}+4\sqrt{2}+9\sqrt{5}+18}{3}+\dfrac{4+2\sqrt{3}}{1}-\dfrac{\sqrt{3}-1}{6}\)
\(=\dfrac{2\left(13\sqrt{5}+4\sqrt{2}+18\right)+24+12\sqrt{3}-\sqrt{3}+1}{6}\)
\(=\dfrac{26\sqrt{5}+4\sqrt{2}+36+25+11\sqrt{3}}{6}\)
\(=\dfrac{61+11\sqrt{3}+26\sqrt{5}+4\sqrt{2}}{6}\)
1: \(\dfrac{12}{5\sqrt{6}}=\dfrac{12\sqrt{6}}{30}=\dfrac{2\sqrt{6}}{5}\)
2: \(\dfrac{3}{2+\sqrt{6}}=\dfrac{-6+3\sqrt{6}}{2}\)
1) \(5\sqrt{8}-\dfrac{7}{2}\sqrt{72}+6\sqrt{\dfrac{1}{2}}\\ =5.\sqrt{4^2.\dfrac{1}{2}}-\dfrac{7}{2}.\sqrt{12^2.\dfrac{1}{2}}+6.\sqrt{\dfrac{1}{2}}=\left(5.4+\dfrac{7}{2}.12+6\right)\sqrt{\dfrac{1}{2}}\\ =68\sqrt{\dfrac{1}{2}}\)
2) \(\dfrac{6}{\sqrt{5}-1}=\dfrac{6.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{6\left(\sqrt{5}+1\right)}{5-1}\\ =\dfrac{6\left(\sqrt{5}+1\right)}{4}=\dfrac{3.\left(\sqrt{5+1}\right)}{2}\)
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
a: \(=5\cdot5\sqrt{3}-\dfrac{1}{3}\cdot3\sqrt{3}=24\sqrt{3}\)
b: \(=\dfrac{12\left(3+\sqrt{5}\right)}{4}=9+3\sqrt{5}\)
c: \(=3-\sqrt{5}+\sqrt{5}=3\)