\(\frac{\sqrt{2}}{\sqrt{ }\sqrt{2+1}-1}-\frac{\sqrt{2}}{\sqrt{ }\sqrt{2+1}+1}\)
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Những câu hỏi liên quan
4 tháng 10 2019
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
28 tháng 11 2020
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n-1}\right)}\)\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n-1}\)
\(A=\sqrt{n}-\sqrt{1}\)
\(B=\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}-\sqrt{2}\right)\left(\sqrt{1}+\sqrt{2}\right)}+\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{\sqrt{24}+\sqrt{25}}{\left(\sqrt{24}-\sqrt{25}\right)\left(\sqrt{24}+\sqrt{25}\right)}\)
\(B=-\left(\sqrt{1}+\sqrt{2}\right)-\left(\sqrt{2}+\sqrt{3}\right)-...-\sqrt{24}+\sqrt{25}\)
\(B=-1-2\sqrt{2}-2\sqrt{3}-...-\sqrt{24}-5\)
\(B=-1-2\sqrt{2}-2\sqrt{3}-...-\sqrt{24}-5\)
\(B=-6-2\sqrt{2}-2\sqrt{3}-...-2\sqrt{24}\)
I
29 tháng 11 2020
ta có \(\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}=\frac{\sqrt{1}-\sqrt{2}}{1-2}=\sqrt{1}-\sqrt{2}\)
mấy cái kia cũng thế a
\(=>A=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-2\right)+...+\left(\sqrt{n}-\sqrt{n-1}\right)\)=>A= căn n -1
18 tháng 11 2015
Nhân cả tử và mẫu với \(\sqrt{3}\)
\(P=\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}=\frac{\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\frac{\sqrt{2}+1+\sqrt{2}-1}{\sqrt{2}+1-\sqrt{2}+1}=\frac{2\sqrt{2}}{2}=\sqrt{2}\)
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
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.........+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2-1}{\sqrt{1}+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+........+\frac{2018-2017}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+......+\)
\(\frac{\left(\sqrt{2018}-\sqrt{2017}\right)\left(\sqrt{2018}+\sqrt{2017}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+........+\left(\sqrt{2018}-\sqrt{2017}\right)\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+......+\sqrt{2018}-\sqrt{2017}\)
\(=-\sqrt{1}+\sqrt{2018}=\sqrt{2018}-\sqrt{1}\)