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\(\frac{4}{2\sqrt{2}+\sqrt{6}-\sqrt{3}-2}\)
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PT
1
15 tháng 7 2016
\(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
\(=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}=\sqrt{6}+\sqrt{2}+\sqrt{5}\)
28 tháng 2 2017
\(A=\dfrac{2}{2.\sqrt[3]{2}+2+\sqrt[3]{2^2}}=\dfrac{2}{\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2}\)
\(A=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\left[\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2\right]}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)^3-\left(\sqrt{2}\right)^3}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{2-2\sqrt{2}}\)
\(A=\dfrac{2\left[.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\right].\left(1+\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\left(\sqrt{2}+1\right)\left(\sqrt{2}-\sqrt[3]{2}\right)\)
2 tháng 7 2019
\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)
\(=\frac{11-\sqrt{10}}{6}\)
\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)
\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)
PT
7
19 tháng 6 2018
4/ Xét \(x< 8\)
\(\Rightarrow C=\frac{4}{1-\frac{4}{x}}\)
Làm nốt
Xét \(x\ge8\)
\(C=\frac{2\sqrt{x-4}}{1-\frac{4}{x}}=\frac{2x\sqrt{x-4}}{x-4}\)
Đặt \(\sqrt{x-4}=a\)
\(\Rightarrow C=\frac{2a\left(a^2+4\right)}{a^2}\ge8\)
YY
0
TY
0
AT
0
3 tháng 8 2020
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
Vậy..
3 tháng 8 2020
b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
Vậy..