Rút gọn:
\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
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4 tháng 10 2019
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
7 tháng 10 2016
\(A=\frac{\sqrt{3}-1}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}=\frac{\sqrt{3}-1}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}+\frac{\sqrt{3}+1}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\sqrt{3}-1}{1+\frac{\sqrt{4+2\sqrt{3}}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{4-2\sqrt{3}}}{2}}=\frac{\sqrt{3}-1}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}+\frac{\sqrt{3}+1}{1-\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\sqrt{3}-1}{\frac{3+\sqrt{3}}{2}}+\frac{\sqrt{3}+1}{\frac{3-\sqrt{3}}{2}}=\frac{2\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{2\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\frac{2}{\sqrt{3}}\left(\frac{4-2\sqrt{3}+4+2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right)=\frac{2}{\sqrt{3}}.\frac{8}{2}=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\)
HN
0
7 tháng 6 2019
với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
NN
6 tháng 9 2020
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.........+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2-1}{\sqrt{1}+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+........+\frac{2018-2017}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+......+\)
\(\frac{\left(\sqrt{2018}-\sqrt{2017}\right)\left(\sqrt{2018}+\sqrt{2017}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+........+\left(\sqrt{2018}-\sqrt{2017}\right)\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+......+\sqrt{2018}-\sqrt{2017}\)
\(=-\sqrt{1}+\sqrt{2018}=\sqrt{2018}-\sqrt{1}\)
Đề bài:
\(\frac{\sqrt{3}}{\sqrt{\sqrt{3+1}-1}}-\frac{\sqrt{3}}{\sqrt{\sqrt{3+1}-1}}\)
= \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)}{\left(\sqrt{\sqrt{3+1}-1}\right).\left(\sqrt{\sqrt{3+1}+1}\right)}.\frac{\sqrt{3}\left(\sqrt{\sqrt{3+1}+1}\right)}{\left(\sqrt{\sqrt{3+1}-1}\right).\left(\sqrt{\sqrt{3+1}+1}\right)}\)
= \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)-\sqrt{3}\left(\sqrt{\sqrt{3+1}+1}\right)}{\left(\sqrt{\sqrt{3+1}-1}\right).\left(\sqrt{\sqrt{3+1}+1}\right)}\)
= \(\frac{\sqrt{3}.\left[\left(\sqrt{\sqrt{3+1}-1}\right).\left(\sqrt{\sqrt{3+1}+1}\right)\right]}{\left(\sqrt{\sqrt{3+1}}\right)^2-1^2}\)
= \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}-1-\sqrt{\sqrt{3+1}}-1\right)}{\sqrt{3}+1-1}\)
= \(\frac{\sqrt{3}.\left(-2\right)}{\sqrt{3}}\)
= -2
Cái này mới đầy đủ và đúng hơn ( cãi nãy là sai vè thiếu nha ) ^^
\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
=\(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\sqrt{3}+1-1}\)
=\(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}}\)
=2