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10 tháng 8 2015
\(12\sqrt{\frac{4}{3}}-\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{4-6\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{3}-2}\)
\(=12.\frac{2}{\sqrt{3}}-\frac{\left(3+\sqrt{2}\right)\left(8-2\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}\left(4-6\sqrt{2}\right)}{2}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{3-4}\)
\(=8\sqrt{3}-\left(4+2\sqrt{2}\right)-\left(2\sqrt{2}-6\right)+\left(-3-2\sqrt{3}\right)\)
\(=8\sqrt{3}-4-2\sqrt{2}-2\sqrt{2}+6-3-2\sqrt{3}\)
\(=6\sqrt{3}-4\sqrt{2}-1\)
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
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
NT
2
11 tháng 3 2020
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
11 tháng 3 2020
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}\) = \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}\)
= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\) = \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
NT
2
12 tháng 3 2020
=\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
12 tháng 3 2020
Ta có :
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
6 tháng 7 2019
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
6 tháng 7 2019
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
OD
1
NN
19 tháng 10 2020
\(D=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right).\left(1+\sqrt{2}\right)}=\frac{1}{\sqrt{2}+1}\)
\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\)
13 tháng 8 2016
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{\sqrt{2}\left(\sqrt{6}+\sqrt{14}\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
4 tháng 10 2019
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{\sqrt{3}+1-2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{(\sqrt{3})^2-2\sqrt{3}.1+1^2}}{\sqrt{2}(\sqrt{3}-1)}=\frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}(\sqrt{3}-1)}=\frac{(\sqrt{3}-1)}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)