Rut Gon bieu thuc
\(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}\)
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Những câu hỏi liên quan
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
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) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)
MT
9 tháng 1 2016
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
14 tháng 6 2018
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
T
0
18 tháng 7 2017
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
Q
4 tháng 6 2017
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) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt[4]{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt[4]{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt[4]{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt[4]{5-\sqrt{\left(1-\sqrt{5}\right)^2}}\)
\(=\sqrt[4]{5-\left(\sqrt{5}-1\right)}\)
\(=\sqrt[4]{5-\sqrt{5}+1}\)
\(=\sqrt[4]{6-\sqrt{5}}\)
\(\sqrt{3-2\sqrt{2}}=\sqrt{1-2\sqrt{2}+2}=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|\)
\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{6}+3}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}=\left|\sqrt{2}-\sqrt{3}\right|\)
Mà\(1< \sqrt{2};\sqrt{2}< \sqrt{3}\)
\(\Rightarrow\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}\)
\(=\sqrt{3}-1\)
ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}.\)
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}=\sqrt{3}-1\)