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![](https://rs.olm.vn/images/avt/0.png?1311)
\(B^3=4+3\sqrt[3]{\left(2+\sqrt[]{5}\right)\left(2-\sqrt[]{5}\right)}\left(\sqrt[3]{2+\sqrt[]{5}}+\sqrt[3]{2-\sqrt[]{5}}\right)\)
\(\Leftrightarrow B^3=4-3B\)
\(\Leftrightarrow B^3+3B-4=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+4\right)=0\)
\(\Leftrightarrow B=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
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\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+3+\sqrt{5}+2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6+2\cdot\sqrt{4}\)
=10
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có: \(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)
\(=\left(\dfrac{2\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-\sqrt{5}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)
\(=\left(2-\sqrt{5}\right)\cdot\dfrac{1}{2-\sqrt{5}}\)
=1
d) Ta có: \(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\dfrac{5}{\sqrt{5}}+\dfrac{1}{3}\cdot3\sqrt{5}+\left|2-\sqrt{5}\right|\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-2\)
\(=3\sqrt{5}-2\)
thực hiện phép tính:\(\sqrt{\left(5-\sqrt{24}^{ }\right)^2}\)- \(\sqrt{\left(5+\sqrt{24}\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\\ =\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|\\ =5-\sqrt{24}-5-\sqrt{24}\\ =-2\sqrt{24}=-4\sqrt{6}\)
`\sqrt((5-\sqrt24)^2) - \sqrt((5+\sqrt24)^2)`
`=|5-\sqrt24|-|5+\sqrt24|`
`=5-\sqrt24-5-\sqrt24`
`=-2\sqrt24`
`=-4\sqrt6`
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Gọi biểu thức trên là $A$
\(A^2=8+2\sqrt{(4+\sqrt{10+2\sqrt{5}})(4-\sqrt{10+2\sqrt{5}})}\)
\(=8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{(\sqrt{5}-1)^2}=8+2|\sqrt{5}-1|=6+2\sqrt{5}=(\sqrt{5}+1)^2\)
$\Rightarrow A=\sqrt{5}+1$ (do $A>0$)
b) 2 2 - 5 - 5 2 + 5
= 2 2 + 5 - 5 2 - 5 2 - 5 2 + 5 = 6 - 7 5