Tính.
`a, sqrt(9-3sqrt(5)` `-sqrt(9+3sqrt(5)`
`b, (sqrt(10) – sqrt(2))“sqrt(4+sqrt(6-2sqrt(5))`
Em cảm ơn trước ạ!
Tính.
`a, sqrt(9-3sqrt(5)` `-sqrt(9+3sqrt(5)`
`b, (sqrt(10) – sqrt(2))“sqrt(4+sqrt(6-2sqrt(5))`
Em cảm ơn trước ạ!
a,
$A=\sqrt{9-3\sqrt5}-\sqrt{9+3\sqrt5}$
$A^2=9-3\sqrt5+9+3\sqrt5-2\sqrt{(9-3\sqrt5)(9+3\sqrt5)}$
$=18-2\sqrt{81-45}$
$=18-2.6$
$=6$
$81>45\to \sqrt{81}>\sqrt{45}\to 9>3\sqrt5>9-3\sqrt5>0$
Ta có: $-3\sqrt5<0<3\sqrt5$
$\to 9-3\sqrt5<9+3\sqrt5$
Khai phương 2 vế:
$\sqrt{9-3\sqrt5}<\sqrt{9+3\sqrt5}$
$\to A<0$
Vậy $A=-\sqrt6$
b,
$(\sqrt{10}-\sqrt2).\sqrt{4+\sqrt{(\sqrt5-1)^2}}$
$=(\sqrt{10}-\sqrt2).\sqrt{4+\sqrt5-1}$
$=(\sqrt{10}-\sqrt2).\sqrt{3+\sqrt5}$
$=(\sqrt5-1).\sqrt2.\sqrt{3+\sqrt5}$
$=(\sqrt5-1).\sqrt{6+2\sqrt5}$
$=(\sqrt5-1).\sqrt{(\sqrt5+1)^2}$
$=(\sqrt5-1)(\sqrt5+1)$
$=5-1$
$=4$
Đáp án:
`a)\sqrt{9-3sqrt5}-sqrt{9+3sqrt5}`
`=sqrt{(18-6sqrt5)/2}-sqrt{(18+6sqrt5)/2}`
`=sqrt{(15-2.sqrt{45}+3)/2}-sqrt{(15+2sqrt{45}+3)/2}`
`=sqrt{(sqrt{15}-sqrt3)^2/2}-sqrt{(sqrt{15}+sqrt3)^2/2}`
`=(sqrt{15}-sqrt3)/2-(sqrt{15}+sqrt3)/sqrt2`
`=(sqrt{15}-sqrt3-sqrt{15}-sqrt3)/sqrt2`
`=(-2sqrt3)/sqrt2`
`=-sqrt6`
`b)(sqrt{10}-sqrt2)sqrt{4+sqrt{6-2sqrt5}}`
`=sqrt2(sqrt5-1)sqrt{4+sqrt{5-2sqrt5+1}}`
`=sqrt2(sqrt5-1)sqrt{4+sqrt{(sqrt5-1)^2}}`
`=sqrt2(sqrt5-1)sqrt{4+sqrt5-1}`
`=sqrt2(sqrt5-1)sqrt{3+sqrt5}`
`=(sqrt5-1)sqrt{2.3+2sqrt5}`
`=(sqrt5-1)sqrt{6+2sqrt5}`
`=(sqrt5-1)sqrt{5+2sqrt5+1}`
`=(sqrt5-1)sqrt{(sqrt5+1)^2}`
`=(sqrt5-1)(sqrt5+1)`
`=(sqrt5)^2-1`
`=5-1=4`.