E=\(\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}-\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}}\)
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MN
8 tháng 8 2020
c) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
d) Đặt \(D=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(\Leftrightarrow D^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(\Leftrightarrow D^2=4+2\sqrt{4-3}\)
\(\Leftrightarrow D^2=6\)
\(\Leftrightarrow D=\sqrt{6}\) (Vì D > 0)
e) \(E=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(\Leftrightarrow E^2=\frac{3-\sqrt{5}}{3+\sqrt{5}}+\frac{3+\sqrt{5}}{3-\sqrt{5}}-2\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}\cdot\frac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(\Leftrightarrow E^2=\frac{9-6\sqrt{5}+5+9+6\sqrt{5}+5}{9-5}-2\sqrt{1}\)
\(\Leftrightarrow E^2=7-2=5\)
\(\Leftrightarrow E=\sqrt{5}\) (Vì E >0)
f) \(\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}\)
\(=\frac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}\)
\(=\frac{2\sqrt{5}}{4}\cdot\frac{1}{\sqrt{5}}\)
\(=\frac{1}{2}\)
NV
16 tháng 7 2016
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/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 9 2019
\(A=\sqrt{3}+\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\sqrt{3}+\sqrt{3}\left(\sqrt{3}-1\right)=3\)
\(B=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\sqrt{3}-2-\sqrt{3}=-2\)
\(C=\frac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}}+\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\sqrt{5}-\sqrt{3}\)
\(C=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
\(D=\frac{2}{\left|2-\sqrt{5}\right|}-\frac{2}{\left|2+\sqrt{5}\right|}=\frac{2}{\sqrt{5}-2}-\frac{2}{\sqrt{5}+2}=\frac{2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}-\frac{2\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(D=2\sqrt{5}+4-2\sqrt{5}+4=8\)
\(E=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}-\sqrt{2}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2019
\(A=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=8-\frac{3+\sqrt{5}}{2}=\frac{13-\sqrt{5}}{2}\)
\(B=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{8+2\sqrt{15}+8-2\sqrt{15}}{2}=8\)
19 tháng 6 2017
Phần d mình sửa lại đề nha : \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{4}-4}\)