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\(A=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{\frac{9-2.3\sqrt{3}+3}{2}}=\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{\sqrt{2}}.\sqrt{\left(3-\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=5\sqrt{2}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{12}\)
\(=-15\sqrt{2}\)
@Nguyễn Thị Thu Sương :
\(\frac{\sqrt{3+\sqrt{15}}}{\sqrt{2}}=\sqrt{\frac{3+\sqrt{15}}{2}}\)
\(=\sqrt{\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{5-3}}\)
\(=\sqrt{\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}}\)
\(=\sqrt{\frac{\sqrt{3}}{\sqrt{5}-\sqrt{3}}}\)
a) \(\left(\sqrt{12}-\sqrt{27}+\sqrt{3}\right):\sqrt{3}\)
\(=\left(2\sqrt{3}-3\sqrt{3}+\sqrt{3}\right):\sqrt{3}\)
\(=\sqrt{3}\left(2-3+1\right):\sqrt{3}\)
\(=0:\sqrt{3}=0\)
b) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
\(=\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
\(=\frac{5\sqrt{3}}{\sqrt{3}\cdot\sqrt{5}}+\frac{3\sqrt{5}}{\sqrt{3}\cdot\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
=\(\sqrt{4}.\sqrt{3}+2\sqrt{9}.\sqrt{3}+3\sqrt{25}.\sqrt{3}-9\sqrt{16}.\sqrt{3}\)
=\(2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
=\(\left(2+6+15-36\right)\sqrt{3}\)
=\(-13\sqrt{3}\)
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) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
