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23 tháng 10 2019

Ta có:

 \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(\sqrt{2}A=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(\sqrt{2}A=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\sqrt{7}-1-\sqrt{7}-1=-2\)

\(\Rightarrow A=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

7 tháng 7 2017

\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

   \(=\sqrt{\frac{2\left(4+\sqrt{7}\right)}{2}}-\sqrt{\frac{2\left(4-\sqrt{7}\right)}{2}}\)

   \(=\sqrt{\frac{8+2\sqrt{7}}{2}}-\sqrt{\frac{8-2\sqrt{7}}{2}}\)

   \(=\sqrt{\frac{7+2\sqrt{7}+1}{2}}-\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)

   \(=\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}\)

   \(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)

   \(=\frac{|\sqrt{7}+1|}{\sqrt{2}}-\frac{|\sqrt{7}-1|}{\sqrt{2}}\)

   \(=\frac{\sqrt{7}+1}{\sqrt{2}}-\frac{\sqrt{7}-1}{\sqrt{2}}\)

   \(=\frac{2}{\sqrt{2}}\)

7 tháng 7 2017

cho mình hỏi tại sao chia 2 vậy?

11 tháng 6 2018

\(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}\right)^2-2^2}-\frac{5\left(4-\sqrt{7}\right)}{4^2-\left(\sqrt{7}\right)^2}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{8}\)

\(=\frac{12\left(\sqrt{7}-5\right)}{24}-\frac{6\left(6+2\sqrt{7}\right)}{24}+\frac{8\left(6\sqrt{7}+12\right)}{24}-\frac{3\left(20-5\sqrt{7}\right)}{24}\)

\(=\frac{12\sqrt{7}-60-36-12\sqrt{7}+48\sqrt{7}+96-60+15\sqrt{7}}{24}\)

\(=\frac{-60+63\sqrt{7}}{24}\)

21 tháng 5 2023

a)

\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)

\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)

\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )

\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)

b)

\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)

\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )

\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)