\(\sqrt{2-\sqrt{ }2\cdot\sqrt{ }5-2}\) - \(\sqrt{2+2\sqrt{ }5-2}\)
Rut gon
K
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19 tháng 1 2019
\(P=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}+\sqrt{3+\sqrt{5}}\right)}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}\)
\(=\dfrac{3\sqrt{2}+\sqrt{10}}{4+\sqrt{6+2\sqrt{5}}}+\dfrac{3\sqrt{2}-\sqrt{10}}{4-\sqrt{6-2\sqrt{5}}}\)
\(=\dfrac{3\sqrt{2}+\sqrt{10}}{5+\sqrt{5}}+\dfrac{3\sqrt{2}-\sqrt{10}}{5-\sqrt{5}}\)
\(=\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(5-\sqrt{5}\right)+\left(3\sqrt{2}-\sqrt{10}\right)\left(5+\sqrt{5}\right)}{20}\)
\(=\dfrac{15\sqrt{2}-3\sqrt{10}+5\sqrt{10}-5\sqrt{2}+15\sqrt{2}+3\sqrt{10}-5\sqrt{10}-5\sqrt{2}}{20}\)
\(=\dfrac{30\sqrt{2}-10\sqrt{2}}{20}=\dfrac{20\sqrt{2}}{20}=\sqrt{2}\)
\(\)
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: \(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}.\)
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}=\sqrt{3}-1\)
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) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)
26 tháng 5 2021
\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)
\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
2 tháng 10 2017
ĐKXĐ:\(x\ge0,x\ne4\)\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}\)=\(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)=\(\frac{3\sqrt{x}}{\sqrt{x}+2}\)
MT
9 tháng 1 2016
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)