T
3
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
TT
0
NH
4
10 tháng 9 2016
\(M>0\Leftrightarrow M^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}..\)
\(M^2=8-2.\sqrt{16-7}=8-6=3\)
\(M=\sqrt{3}.\)
7 tháng 8 2018
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
30 tháng 6 2017
a.\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right).\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(=\left(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\right).\left(\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\right)\)
\(=\left(\sqrt{3}+1-\sqrt{3}+1\right)\left(\sqrt{3}-1+\sqrt{3}+1\right)\)
\(=2.2\sqrt{3}=4\sqrt{3}\)
b.\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=\left[\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\right]^2\)
\(=\left(\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\right)^2\)
\(=\left(\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}\right)^2=\left(\sqrt{2}\right)^2=2\)
c.\(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}=\sqrt{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
\(=\sqrt{9+4\sqrt{3}+4}\)
= \(\sqrt{\left(\sqrt{3}+2\right)^2}\)
=\(\sqrt{3}+2\)
\(\sqrt{9+4+4\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\sqrt{3}+2\)