`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`

`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`

`= sqrt3/sqrt 7`

`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`

`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`

`= sqrt5`

`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`

`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`

`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`

`((5-2 sqrt 5)/(2-sqrt5) - 2)((5+3 sqrt 5)/(3+sqrt 5) - 2)`

`= (5-2sqrt 5 - 4 + 2 sqrt 5)/(2 -sqrt 5) . (5+3 sqrt 5 - 6 - 2 sqrt 5)/(3 + sqrt 5)`

`= 1/(2-sqrt5) . (-1 + sqrt 5)/(3 + sqrt 5)`

`= (sqrt 5 - 1)/((sqrt 5 + 3)(2 - sqrt 5))`

`b, (sqrt(15-10sqrt2) - sqrt 10)/(sqrt 5)`

`= (sqrt(10 - 2 . sqrt 5.sqrt 10 + 5) - sqrt 10)/(sqrt 5)`

`= (sqrt 10 - sqrt 5 - sqrt 10)/(sqrt5)`

`= -1`

ghê dữ zậy cho tớ xin in4 với :>

`a + b = |a| - |b|`

`a + b + |b| = |a|`

`@ a >=0 => a >=a`.

`<=> b <=0`.

`@ a < 0 => b + |b| = |a| - a`

`=> a = b`.

`a, 9 xx (1/3)^2 + (1/7)^0 - 1`

`= 9 xx 1/9 + 1 -1`

`= 1`

`b, 7(5^2-5) + (-1)^2`

`= 7 . 20 + 1`

`= 141`

`a, A = sqrt((2 sqrt 3 - 3 sqrt 2)^2) + sqrt(13-4sqrt3) - sqrt(22+12sqrt2)`

`= | 2 sqrt 3 - 3 sqrt 2| + sqrt(12-4sqrt3 + 1) - sqrt(18 + 12 sqrt 2 + 4)`

`= 3 sqrt 2 - 2 sqrt 3 + 2 sqrt 3 - 1 - (3 sqrt 2 + 2)`.

`= 3 sqrt 2 - 2 sqrt 3 + 2 sqrt 3 - 3 sqrt 2 - 3`

`= -3`

`b, B = sqrt(17-6sqrt(2+sqrt(9+4sqrt2)))`

`= sqrt(17-6sqrt(2+sqrt((2 sqrt 2+1)^2)))`

`= sqrt(17 - 6sqrt(3 + 2 sqrt 2))`

`= sqrt(17 - 6 sqrt((sqrt 2 + 1)^2))`

`= sqrt(17-6(sqrt2+1))`

`= sqrt(11 - 6 sqrt 2)`

`= sqrt(9 - 6 sqrt 2 + 2)`

`= 3 - sqrt 2`.

`a, sqrt((sqrt 15 - 4)^2)`

`= | sqrt 15 - 4|`

`= 4 - sqrt 15`.

`b, sqrt(7-2 sqrt 6)`

`= sqrt(6 - 2 sqrt 6 + 1)`

`= sqrt((sqrt 6-1)^2)`

`= sqrt 6 - 1`