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30 tháng 12 2020

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

vậy bình phương

 \(A^2=4-\sqrt{7}+2.\sqrt{4-\sqrt{7}.4+\sqrt{7}}+4+\sqrt{7}\)

\(A^2=8+2\sqrt{16}=8+8=16\)

Do vậy dễ thấy \(A=\sqrt{16}=4\)

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

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

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

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

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

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

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

23 tháng 10 2019

Ta có:

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

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

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

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

15 tháng 7 2019

TL:

\(\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\) 

\(=\frac{8-3\sqrt{7}-8-3\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)

\(=\frac{-6\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)

15 tháng 7 2019

Cho   \(A=\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)

CACH  1  : \(\Rightarrow A\sqrt{2}=\sqrt{16-6\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)

\(\Rightarrow A\sqrt{2}=\sqrt{9-2.3.\sqrt{7}+7}-\sqrt{9+2.3.\sqrt{7}+7}\)

\(\Rightarrow A\sqrt{2}=\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(3+\sqrt{7}\right)^2}\)

\(\Rightarrow A\sqrt{2}=|3-\sqrt{7}|-|3+\sqrt{7}|\)

\(\Rightarrow A\sqrt{2}=3-\sqrt{7}-3-\sqrt{7}=-2\sqrt{7}=-\sqrt{28}\)

\(\Rightarrow A=-\sqrt{14}\)

CACH   2  :   \(A^2=8-3\sqrt{7}+8+3\sqrt{7}-2.\sqrt{8^2-\left(3\sqrt{7}\right)^2}\)

\(\Rightarrow A^2=16-2\sqrt{64-63}=16-2=14\)

\(\Rightarrow A=\sqrt{14}\) hoặc  \(A=-\sqrt{14}\)

Mà  \(8+3\sqrt{7}>8-3\sqrt{7}\) \(\Rightarrow\sqrt{8+3\sqrt{7}}>\sqrt{8-3\sqrt{7}}\)

Vây  A  âm  \(\Rightarrow A=-\sqrt{14}\)

6 tháng 8 2019

a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)

       = \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)

       = \(\frac{-2\sqrt{6}}{2}\)

       = \(-\sqrt{6}\)

17 tháng 12 2016

a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)

b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)

d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)

C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)