Tính: \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
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ND
1
30 tháng 6 2021
$(4+\sqrt{15})(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}.(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=(\sqrt{10}-\sqrt6)\sqrt{4+\sqrt{15}}\sqrt{16-15}$
$=\sqrt2(\sqrt5-\sqrt3)\sqrt{4+\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{8+2\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{5+2\sqrt{5}.\sqrt3+3}$
$=(\sqrt5-\sqrt3)\sqrt{(\sqrt5+\sqrt3)^2}$
$=(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)=5-3=2$
H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
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) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
NN
1
16 tháng 9 2021
\(=\sqrt{4+\sqrt{15}}\left(\sqrt{4+\sqrt{15}}\cdot\sqrt{4-\sqrt{15}}\right)\left(\sqrt{10}-\sqrt{6}\right)\\ =\sqrt{4+\sqrt{15}}\left(16-15\right)\left(\sqrt{10}-\sqrt{6}\right)\\ =\sqrt{2\left(4+\sqrt{15}\right)}\left(\sqrt{5}-\sqrt{3}\right)\\ =\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\\ =\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
SY
1
G
8 tháng 7 2019
\(A=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\)
\(A=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(A=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
16 tháng 7 2016
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\left(\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\right).\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}\)
\(=\sqrt{4^2-15}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)
LG
0
19 tháng 7 2020
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: \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\cdot\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left[4^2-\left(\sqrt{15}\right)^2\right]\)
\(=2\cdot\left[16-15\right]=2\cdot1=2\)
TH
rút gọn
C=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
2
20 tháng 5 2022
\(C=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
2
20 tháng 5 2022
`C=(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}`
`C=(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10})\sqrt{4-\sqrt{15}}`
`C=(\sqrt{10}+\sqrt{6})\sqrt{4-\sqrt{15}}`
`C=\sqrt{(\sqrt{10}+\sqrt{6})^2 .(4-\sqrt{15})}`
`C=\sqrt{(10+6+2\sqrt{60})(4-\sqrt{15})}`
`C=\sqrt{(16+4\sqrt{15})(4-\sqrt{15})}`
`C=\sqrt{64-16\sqrt{15}+16\sqrt{15}-60}`
`C=\sqrt{4}=2`
TK
4
21 tháng 8 2017
\(\sqrt{4+\sqrt{15}}.\)\(\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}.\left(\sqrt{10}-\sqrt{6}\right)\)
=\(1.\sqrt{4+\sqrt{15}}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
=\(\sqrt{8+2\sqrt{15}}.\left(\sqrt{5}-\sqrt{3}\right)=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
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