TRục căn thức ở mẫu : \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\)
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1 tháng 8 2015
a; \(=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{3+2}=\frac{\sqrt[3]{3}+\sqrt[3]{2}}{5}\)
b; tương tự
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..
NV
1
24 tháng 12 2018
\(\hept{\begin{cases}\sqrt[3]{3}=a\\\sqrt[3]{4}=b\end{cases}}\)
\(\Rightarrow b^3-a^3=1\)
\(\Leftrightarrow-b^2-ab=a^2+\frac{1}{a-b}\)
Ta cần trục cái:
\(\frac{1}{a^2-ab-b^2}=\frac{1}{a^2+a^2+\frac{1}{a-b}}=\frac{a-b}{2a^3-2a^2b+1}\)
\(=\frac{\sqrt[3]{3}-\sqrt[3]{4}}{7-2\sqrt[3]{36}}=\frac{\left(\sqrt[3]{3}-\sqrt[3]{4}\right)\left(49+14\sqrt[3]{36}+24\sqrt[3]{6}\right)}{55}=\frac{\sqrt[3]{3}-7\sqrt[3]{4}-4\sqrt[3]{18}}{55}\)
PT
1
15 tháng 7 2016
\(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
\(=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}=\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}=\sqrt{6}+\sqrt{2}+\sqrt{5}\)
2 tháng 7 2019
\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)
\(=\frac{11-\sqrt{10}}{6}\)
\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)
\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)
CS
0
28 tháng 2 2017
\(A=\dfrac{2}{2.\sqrt[3]{2}+2+\sqrt[3]{2^2}}=\dfrac{2}{\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2}\)
\(A=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\left[\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2\right]}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)^3-\left(\sqrt{2}\right)^3}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{2-2\sqrt{2}}\)
\(A=\dfrac{2\left[.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\right].\left(1+\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\left(\sqrt{2}+1\right)\left(\sqrt{2}-\sqrt[3]{2}\right)\)
Ta có: \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}=\)\(\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{2}+\sqrt[3]{3}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{3}\right)^3}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{5}\)