\(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)
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29 tháng 8 2021
2: \(\dfrac{\sqrt{108}}{\sqrt{3}}=6\)
13: \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
14: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
=2
HP
29 tháng 8 2021
12.
\(\dfrac{\sqrt{108}}{\sqrt{3}}=\dfrac{\sqrt{36}.\sqrt{3}}{\sqrt{3}}=\sqrt{36}=6\)
13.
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{5}\right|-\left|2\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
26 tháng 6 2021
`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`
`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`
`<=>A^2=8+2sqrt{6-2sqrt5}`
`<=>A^2=8+2sqrt{(sqrt5-1)^2}`
`<=>A^2=8+2(sqrt5-1)`
`<=>A^2=6+2sqrt5=(sqrt5+1)^2`
`<=>A=sqrt5+1(do \ A>0)`
`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`
Vì `35+12sqrt6>35-12sqrt6`
`=>B>0`
`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`
`<=>B^2=70-2sqrt{361}`
`<=>B^2=70-2sqrt{19^2}=70-38=32`
`<=>B=sqrt{32}=4sqrt2(do \ B>0)`
`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`
`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`
`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`
`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`
`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`
`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`
`=(sqrt5+sqrt3)(sqrt5-sqrt3)`
`=5-3=2`
HA
1
29 tháng 10 2021
\(=\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}+\sqrt{6}\\ =\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{3}{2}}+\sqrt{6}\\ =2\sqrt{\dfrac{3}{2}}+\sqrt{6}=\sqrt{6}+\sqrt{6}=2\sqrt{6}\)
MG
1
H9
HT.Phong (9A5)
CTVHS
20 tháng 6 2023
\(\left(\sqrt{\text{1}0}-\sqrt{6}\right)\cdot\left(4+\sqrt{15}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(10+6+2\sqrt{60}\right)\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(16+4\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)
\(=\sqrt{64-16\sqrt{15}+16\sqrt{15}-60}\)
\(=\sqrt{4}=2\)
NM
1
S
29 tháng 12 2020
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=\left(4+\sqrt{15}\right)\left(4-2\sqrt{15}\right).2\)
\(=\left(4^2-15\right).2\)
\(=2\left(ĐPCM\right)\)
ND
1
30 tháng 6 2021
$(4+\sqrt{15})(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}.(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=(\sqrt{10}-\sqrt6)\sqrt{4+\sqrt{15}}\sqrt{16-15}$
$=\sqrt2(\sqrt5-\sqrt3)\sqrt{4+\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{8+2\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{5+2\sqrt{5}.\sqrt3+3}$
$=(\sqrt5-\sqrt3)\sqrt{(\sqrt5+\sqrt3)^2}$
$=(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)=5-3=2$
22 tháng 7 2017
a) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1+\sqrt{2}\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
= \(\dfrac{8+2\sqrt{5}}{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
=\(\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}.\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\dfrac{\left(5-3\right)^2}{2}=2\)
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) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
\(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)
\(=\frac{\sqrt{2}.\sqrt{4-\sqrt{15}}}{\sqrt{2}}-\frac{\sqrt{2}.\sqrt{4+\sqrt{15}}}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\sqrt{2.\left(4-\sqrt{15}\right)}}{\sqrt{2}}-\frac{\sqrt{2.\left(4+\sqrt{15}\right)}}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}-\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}}{\sqrt{2}}-\frac{\sqrt{5+2.\sqrt{5}.\sqrt{3}+3}}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}-\frac{\left|\sqrt{5}+\sqrt{3}\right|}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\sqrt{6}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\sqrt{6}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}-\frac{2\sqrt{6}}{2}\)
\(=\frac{2\sqrt{6}}{2}-\frac{2\sqrt{6}}{2}\)
\(=0\)
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