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Tính
a) Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)-\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{4-\sqrt{15}}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\sqrt{5}-\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{3}+\sqrt{5}-\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{5}+\sqrt{3}\)
\(=2\sqrt{3}\)
c) Ta có: \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\cdot\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left[4^2-\left(\sqrt{15}\right)^2\right]\)
\(=2\cdot\left[16-15\right]=2\cdot1=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
$(4+\sqrt{15})(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}.(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=(\sqrt{10}-\sqrt6)\sqrt{4+\sqrt{15}}\sqrt{16-15}$
$=\sqrt2(\sqrt5-\sqrt3)\sqrt{4+\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{8+2\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{5+2\sqrt{5}.\sqrt3+3}$
$=(\sqrt5-\sqrt3)\sqrt{(\sqrt5+\sqrt3)^2}$
$=(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)=5-3=2$
![](https://rs.olm.vn/images/avt/0.png?1311)
B=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
=\(\left(4\sqrt{10}+5\sqrt{6}-4\sqrt{6}-3\sqrt{10}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
=\(\left(\sqrt{6}+\sqrt{10}\right)\left(\sqrt{4-\sqrt{15}}\right)\)=\(\left(\sqrt{24-6\sqrt{15}}\right)+\left(\sqrt{40-10\sqrt{15}}\right)\)
=\(\sqrt{15}-3+5-\sqrt{15}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)
\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)
\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)
\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)
\(=\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}-3\right)^2}\)
\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)
\(=7-\sqrt{21}+\sqrt{21}-3\)
\(=4\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}.\sqrt{3}+3}=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2=vp\)Vậy , đẳng thức được chứng minh .
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\)
\(A=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(A=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(4+căn 15) nhân (căn 10 -căn 6) nhân (căn tất cả của 4-căn 15)
= (4+ căn 15) (căn 5 - căn 3) (căn 2) (căn (4- căn 15))
= (4+ căn 15) (căn 5 - căn 3) (căn (8- 2 căn 15))
= (4+ căn 15) (căn 5 - căn 3) (căn 5 - căn 3)
= (4+ căn 15) (5 + 3 - 2 căn 15)
= (4+ căn 15) (4.2 - 2 . căn 15)
= 2.(4+ căn 15) (4- căn 15)
= 2. (16-15)
=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)
\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)
https://hoc24.vn/hoi-dap/question/405366.html
\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
= 5 - 3
= 2
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\frac{1}{2}\left(8+2\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\frac{1}{2}\left(\sqrt{5}+\sqrt{3}\right)^2\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\frac{1}{2}\left(5-3\right)^2=2\)