rut gon bieu thuc a) \(\sqrt{15+2\sqrt{5}-\sqrt[]{21-4\sqrt{5}}}\)
b)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
MT
9 tháng 1 2016
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
8 tháng 8 2016
1) \(\frac{\sqrt{6-2\sqrt{5}}}{2-2\sqrt{5}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{2\left(1-\sqrt{5}\right)}=\frac{\sqrt{5}-1}{2\left(1-\sqrt{5}\right)}=-\frac{1}{2}\)
2) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)
TQ
2
14 tháng 10 2017
\(\sqrt{3-2\sqrt{2}}=\sqrt{1-2\sqrt{2}+2}=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|\)
\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{6}+3}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}=\left|\sqrt{2}-\sqrt{3}\right|\)
Mà\(1< \sqrt{2};\sqrt{2}< \sqrt{3}\)
\(\Rightarrow\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}\)
\(=\sqrt{3}-1\)
14 tháng 10 2017
ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}.\)
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}=\sqrt{3}-1\)
MM
0
a) \(\sqrt{15+2\sqrt{5}-\sqrt{21-4\sqrt{5}}}\)
\(=\sqrt{15+2\sqrt{5}-\sqrt{\left(1-2\sqrt{5}\right)^2}}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-2\sqrt{5}+1}\)
\(=\sqrt{16}\)
\(=4\)
b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt[4]{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt[4]{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt[4]{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt[4]{5-\sqrt{\left(1-\sqrt{5}\right)^2}}\)
\(=\sqrt[4]{5-\left(\sqrt{5}-1\right)}\)
\(=\sqrt[4]{5-\sqrt{5}+1}\)
\(=\sqrt[4]{6-\sqrt{5}}\)