(\(\sqrt{10}-\sqrt{6}\)).(4+\(\sqrt{15}\)).\(\sqrt{4-\sqrt{ }15}\)
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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`
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.
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{5}\right|-\left|2\sqrt{5}-\sqrt{3}\right|\)
\(=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}\)
\(=-\sqrt{5}\)
NM
1
S
29 tháng 12 2020
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=\left(4+\sqrt{15}\right)\left(4-2\sqrt{15}\right).2\)
\(=\left(4^2-15\right).2\)
\(=2\left(ĐPCM\right)\)
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: \(C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\cdot\left(8-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left[4^2-\left(\sqrt{15}\right)^2\right]\)
\(=2\cdot\left[16-15\right]=2\cdot1=2\)
ND
1
30 tháng 6 2021
$(4+\sqrt{15})(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}.(\sqrt{10}-\sqrt6)\sqrt{4-\sqrt{15}}$
$=(\sqrt{10}-\sqrt6)\sqrt{4+\sqrt{15}}\sqrt{16-15}$
$=\sqrt2(\sqrt5-\sqrt3)\sqrt{4+\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{8+2\sqrt{15}}$
$=(\sqrt5-\sqrt3)\sqrt{5+2\sqrt{5}.\sqrt3+3}$
$=(\sqrt5-\sqrt3)\sqrt{(\sqrt5+\sqrt3)^2}$
$=(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)=5-3=2$
3
21 tháng 8 2021
\(B=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\\ B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ B=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\\ B=2\)
21 tháng 8 2021
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4+\sqrt{15}}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{16-15}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}.\sqrt{5}+\left(\sqrt{5}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)=\left|\sqrt{5}+\sqrt{3}\right|\left(\sqrt{5}-\sqrt{3}\right)=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(\left(\sqrt{\text{1}0}-\sqrt{6}\right)\cdot\left(4+\sqrt{15}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{10}+\sqrt{6}\right)^2\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(10+6+2\sqrt{60}\right)\left(4-\sqrt{15}\right)}\)
\(=\sqrt{\left(16+4\sqrt{15}\right)\left(4-\sqrt{15}\right)}\)
\(=\sqrt{64-16\sqrt{15}+16\sqrt{15}-60}\)
\(=\sqrt{4}=2\)