rút gọn
\(\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
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28 tháng 5 2016
Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
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
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
1 tháng 10 2016
\(D=\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right).\frac{\sqrt{3}-1}{3-\sqrt{3}}=\left(\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{-1}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{-1}\right).\frac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\left(-5-2\sqrt{6}+5-2\sqrt{6}\right).\frac{1}{\sqrt{3}}=\frac{-4\sqrt{6}}{\sqrt{3}}=-4\sqrt{2}\)
\(E=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}=\frac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2}{\sqrt{2}}\)
\(=\sqrt{2}\)
NC
17 tháng 8 2020
Bài làm:
a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)
\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)
\(A=4+2\sqrt{3}+5\sqrt{3}-1\)
\(A=3+7\sqrt{3}\)
b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
NC
17 tháng 8 2020
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)
DD
31 tháng 7 2015
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+...-\frac{1}{\sqrt{2013}-\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{\sqrt{3}+\sqrt{4}}{3-4}+...+\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)
\(=-\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{3}+\sqrt{4}-\left(\sqrt{4}+\sqrt{5}\right)+...+\sqrt{2014}+\sqrt{2015}\)
=\(-\sqrt{2}+\sqrt{2015}\)
đẶT \(A=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(=\sqrt{\frac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{11}}-\sqrt{\frac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{13}}\)
\(=\sqrt{\frac{18-3\sqrt{3}-8\sqrt{3}+4}{11}}-\sqrt{\frac{5\sqrt{3}+6+20+8\sqrt{3}}{13}}\)
\(=\sqrt{\frac{11\left(2-\sqrt{3}\right)}{11}}-\sqrt{\frac{13\left(2+\sqrt{3}\right)}{13}}\)
\(=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
ta có: \(2-\sqrt{3}< 2+\sqrt{3}\Rightarrow\sqrt{2-\sqrt{3}}< \sqrt{2+\sqrt{3}}\)
\(\Rightarrow A< 0\Rightarrow-A>0\)
\(\Rightarrow-A=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(A^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2\)
\(A^2=\left(\sqrt{2+\sqrt{3}}\right)^2-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\left(\sqrt{2-\sqrt{3}}\right)^2\)
\(A^2=\left|2+\sqrt{3}\right|-2\sqrt{4-3}+\left|2-\sqrt{3}\right|\)
\(A^2=2+\sqrt{3}-2+2-\sqrt{3}\)
\(A^2=2\)
\(A=\pm\sqrt{2}\)
mà -A > 0 nên A = \(-\sqrt{2}\)
~~ Học tốt ~~
Ở dòng:
\(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\) còn có thêm cách phân tích
\(\sqrt{2}.A=\sqrt{4-2.\sqrt{3}}-\sqrt{4+2.\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1-\sqrt{3}-1=-2\)
=> \(A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)