Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
=căn 5-1/2*căn 5
=1/2*căn 5
e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)
f:=2+căn 3+căn 2-2-căn 3=căn 2
\(a.\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=2\sqrt{5}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}-2}{1-\sqrt{5}}=\dfrac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\) \(b.\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6.5}+\sqrt{27.6}}=\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{17}\right)}=-\dfrac{\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-2-1}{\sqrt{6}}=-\dfrac{\sqrt{3}}{\sqrt{2}}\)
a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).
Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{4\sqrt{2}-3\sqrt{2}}{3\sqrt{2}-4\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=-1-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-6-\sqrt{6}}{6}\)
\(=\dfrac{2\sqrt{8}-2\sqrt{3}}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\cdot\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
a: \(5\sqrt{2}-8\sqrt{3}+30\sqrt{3}-6\sqrt{3}=5\sqrt{2}+16\sqrt{3}\)
b: \(=14\sqrt{3}-\dfrac{3}{32}\cdot8\sqrt{3}+\dfrac{4}{18}\cdot9\sqrt{3}-\dfrac{1}{10}\cdot10\sqrt{3}\)
\(=14\sqrt{3}-\dfrac{3}{4}\sqrt{3}+2\sqrt{3}-1\sqrt{3}=\dfrac{57}{4}\sqrt{3}\)
c: \(=\dfrac{-1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{22}\cdot11\sqrt{3}+2\sqrt{3}\)
\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{1}{2}\sqrt{3}+2\sqrt{3}=-\dfrac{7}{6}\sqrt{3}\)
d: \(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}-\dfrac{1}{4}\cdot8\sqrt{3}\)
\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}-2\sqrt{3}=\dfrac{5}{3}\sqrt{3}\)
b: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8}{\sqrt{5}-1}\)
\(=2\sqrt{5}-2-2\sqrt{5}\)
=-2
c: \(=\dfrac{\sqrt{4}\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{\sqrt{6}}{2}\)