K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
10 tháng 8 2015
\(12\sqrt{\frac{4}{3}}-\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{4-6\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{3}-2}\)
\(=12.\frac{2}{\sqrt{3}}-\frac{\left(3+\sqrt{2}\right)\left(8-2\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}\left(4-6\sqrt{2}\right)}{2}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{3-4}\)
\(=8\sqrt{3}-\left(4+2\sqrt{2}\right)-\left(2\sqrt{2}-6\right)+\left(-3-2\sqrt{3}\right)\)
\(=8\sqrt{3}-4-2\sqrt{2}-2\sqrt{2}+6-3-2\sqrt{3}\)
\(=6\sqrt{3}-4\sqrt{2}-1\)
6 tháng 1 2017
Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)(nhân lượng liên hiệp nhé)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta có
\(\frac{1}{2\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
NT
3
PL
11 tháng 3 2020
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 ~~
11 tháng 3 2020
Ở dòng:
\(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\) còn có thêm cách phân tích
\(\sqrt{2}.A=\sqrt{4-2.\sqrt{3}}-\sqrt{4+2.\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1-\sqrt{3}-1=-2\)
=> \(A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
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
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
8 tháng 7 2018
\(a.\sqrt{\frac{2-\sqrt{3}}{2}}+\frac{1-\sqrt{3}}{2}\)
\(=\sqrt{\frac{2\left(2-\sqrt{3}\right)}{4}}+\frac{1-\sqrt{3}}{2}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{2}+\frac{1-\sqrt{3}}{2}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}+\frac{1-\sqrt{3}}{2}\)
\(=\frac{\sqrt{3}-1+1-\sqrt{3}}{2}\) ( Vì \(\sqrt{3}-1>0\))
\(=0\)
b) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}+\frac{\sqrt{3}}{3}-\frac{2\left(3-\sqrt{3}\right)}{3^2-\left(\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}-\frac{3-\sqrt{3}}{3}\)
\(=\frac{6-3+\sqrt{3}}{3}\)
\(=\frac{3+\sqrt{3}}{3}=\frac{\sqrt{3}+1}{\sqrt{3}}\)
c) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{2\left(2-\sqrt{3}\right)}{1}+\frac{13\left(1+\sqrt{3}\right)}{13}+2\sqrt{3}\)
\(=4-2\sqrt{3}+1-\sqrt{3}+2\sqrt{3}\)
\(=5-\sqrt{3}\)
TT
0