Tính :A=\(\frac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\frac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
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17 tháng 8 2019
\(H=\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)\(-\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)(cái này cùng dòng với cái phía trên)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{2\sqrt{3}}\)
\(H=\frac{-4}{2\sqrt{3}}\)
\(H=\frac{-2}{\sqrt{3}}\)
\(H=-\frac{2\sqrt{3}}{3}\)
NH
17 tháng 8 2019
Đặt \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(A^2=4+2\sqrt{4+2\sqrt{3}-2\sqrt{3}-3}\)
\(A^2=4+2=6\)
\(A=\sqrt{6}\)
Đặt \(B=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(B^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(B^2=4-2\sqrt{4+2\sqrt{3}-2\sqrt{3}-3}\)
\(B^2=4-2\sqrt{1}=4-2=2\)
\(B=\sqrt{2}\)
Thay vào H
\(\Rightarrow H=\frac{\sqrt{2}}{\sqrt{6}}-\frac{\sqrt{6}}{\sqrt{2}}=\frac{1}{\sqrt{3}}-\sqrt{3}=\frac{1-3}{\sqrt{3}}=\frac{-2}{\sqrt{3}}\)
13 tháng 7 2017
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
5 tháng 8 2019
a) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)
HT
17 tháng 6 2019
a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)
\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)
b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)
\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)
c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)
\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)
d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)
\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)
\(A=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{4-2\sqrt{3}}+4}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{4+2\sqrt{3}}-4}=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{\left(\sqrt{3}-1\right)^2}+4}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{\left(\sqrt{3}+1\right)^2}-4}\)
\(A=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{3}-3}=\frac{\sqrt{2}\left(\sqrt{3}-3\right)^2+\sqrt{2}\left(\sqrt{3}+3\right)^2}{3-9}=\frac{24\sqrt{2}}{-6}=-4\sqrt{2}\)