Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
a)\(\sqrt{\dfrac{4}{9-4\sqrt{5}}}-\sqrt{\dfrac{4}{9+4\sqrt{5}}} \Leftrightarrow \dfrac{\sqrt{4}}{\sqrt{(2-\sqrt{5}})^{2}}-\dfrac{\sqrt{4}}{(2+\sqrt{5})^{2}} \Leftrightarrow \dfrac{2(2+\sqrt{5})}{(\sqrt{5}-2)(2+\sqrt{5})}-\dfrac{2(\sqrt{5}-2)}{(\sqrt{5}-2)(2+\sqrt{5})} \Leftrightarrow \dfrac{4+2\sqrt{5}-(2\sqrt{5}-4)}{4-5} \Leftrightarrow \dfrac{8}{-1} = -8\)b)\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}} =\dfrac{\sqrt{2}\sqrt{8-4\sqrt{3}}}{\sqrt{2}\sqrt{2}} =\dfrac{\sqrt{16-8\sqrt{3}}}{2} =\dfrac{\sqrt{(2-2\sqrt{3})^{2}}}{2} =\dfrac{2\sqrt{3}-2}{2} =\dfrac{2(\sqrt{3}-1)}{2} =\sqrt{3}-1\)c)\(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}} =\sqrt{2}\sqrt{7-4\sqrt{3}}-\sqrt{2}\sqrt{12+6\sqrt{3}} =\sqrt{2}(\sqrt{(4-\sqrt{3})^{2}}-\sqrt{(3+\sqrt{3})^{2}}) =\sqrt{2}((4-\sqrt{3})-(3+\sqrt{3})) =\sqrt{2}(1-2\sqrt{3}) =\sqrt{2}-2\sqrt{6}\)
\(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
mik chỉnh lại đề
\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)
\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{15+2.3.\sqrt{6}}\)\(-\sqrt{10+2.2\sqrt{6}}\)
\(=\sqrt{9+2.3\sqrt{6}+6}\)\(-\sqrt{6+2.\sqrt{6}.2+4}\)
\(=\sqrt{\left(3+\sqrt{6}\right)^2}\)\(-\sqrt{\left(\sqrt{6}+2\right)^2}\)
\(=3+\sqrt{6}\)\(-2\)\(-\sqrt{6}=\left(3-2\right)+\left(\sqrt{6}-\sqrt{6}\right)\)
\(=1+0=1\)
a) \((\sqrt{3}-\sqrt{2}).\sqrt{(\sqrt{3}+\sqrt{2})^2}\)
\(\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right)\)
\(\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)\(=3-2=1\)
b) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
=\(\sqrt{(2+2\sqrt{5})^2}+\sqrt{(\sqrt{5}-2)^2}\)
=\(2+2\sqrt{5}+\sqrt{5}-2\)\(=3\sqrt{5}\)
a/ \(\sqrt{21+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
= \(\sqrt{16+2.4.\sqrt{5}+5}+\sqrt{5-2.2\sqrt{5}+4}\)
= \(\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
= \(4+\sqrt{5}+\sqrt{5}-2=2+2\sqrt{5}\)
b/ \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{10}+\sqrt{15}}{2\left(\sqrt{2}+\sqrt{3}\right)}\) = \(\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(\dfrac{\sqrt{20}-\sqrt{30}+\sqrt{30}-\sqrt{45}}{2\left(2-3\right)}\) = \(\dfrac{\sqrt{20}-\sqrt{45}}{-2}\) = \(\dfrac{2\sqrt{5}-3\sqrt{5}}{-2}\)
= \(\dfrac{-\sqrt{5}}{-2}=\dfrac{\sqrt{5}}{2}\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
a: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)
b: \(=\sqrt{2}-1-\sqrt{2}=-1\)
a)\(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\sqrt{\dfrac{1}{2}\left(16+8\sqrt{3}\right)}-\sqrt{\dfrac{1}{2}\left(16-8\sqrt{3}\right)}\)
\(=\sqrt{\dfrac{1}{2}\left(2+2\sqrt{3}\right)^2}-\sqrt{\dfrac{1}{2}\left(2-2\sqrt{3}\right)^2}\)\(=\sqrt{\dfrac{1}{2}}\left(2+2\sqrt{3}\right)-\sqrt{\dfrac{1}{2}}\left(2\sqrt{3}-2\right)=2\sqrt{2}\)
b)\(=\dfrac{\sqrt{16+2.4\sqrt{5}+5}}{4+\sqrt{5}}.\sqrt{\left(2-\sqrt{5}\right)^2}\)\(=\dfrac{\sqrt{\left(4+\sqrt{5}\right)^2}}{4+\sqrt{5}}\left|2-\sqrt{5}\right|=\sqrt{5}-2\)
a) Ta có: \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)
\(=\sqrt{6}+\sqrt{2}-\sqrt{6}+\sqrt{2}\)
\(=2\sqrt{2}\)
b) Ta có: \(\dfrac{\sqrt{21+8\sqrt{5}}}{4+\sqrt{5}}\cdot\sqrt{9-4\sqrt{5}}\)
\(=\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)\)
=16-5=11