Helppppp meeeeeeeeee

`A = ( \sqrt{( \sqrt{3} + 1 )^2} . ( \sqrt{3} -1 ))/( \sqrt{( \sqrt{5} + 1 )^2} - \sqrt{5} )`

`A = ( | \sqrt{3} + 1 |( \sqrt{3} - 1 ) )/( | \sqrt{5} + 1 | - \sqrt{5} )`

`A = (( \sqrt{3} + 1 )(\sqrt{3} - 1 ))/( \sqrt{5} + 1 - \sqrt{5})`

`A = ( 3 - 1 )/1`

`A = 2`

ĐKXĐ `x > 0 ; x \ne 1`

`B = ( 1/( x - \sqrt{x} ) + 1/( \sqrt{x} - 1 ) ) : ( \sqrt{x} + 1 )/(( \sqrt{x} - 1 )^2)`

`B = ( 1/(\sqrt{x}( \sqrt{x} - 1 )) + ( \sqrt{x} )/(\sqrt{x}( \sqrt{x}- 1))) . (( \sqrt{x} - 1 )^2)/( \sqrt{x} + 1 )`

`B = ( \sqrt{x} + 1 )/(\sqrt{x}( \sqrt{x} - 1 )) . (( \sqrt{x} - 1 )^2)/( \sqrt{x} + 1 )`

`B = ( \sqrt{x} - 1 )/( \sqrt{x})`

`A + B <= 0`

`( \sqrt{x} - 1 )/( \sqrt{x} ) + 2 <= 0`

`( 3\sqrt{x} - 1 )/( \sqrt{x}) <= 0`

`Do  x > 0 => \sqrt{x} > 0`

`3\sqrt{x} - 1 <= 0`

`3\sqrt{x} <= 1`

`\sqrt{x} <= 1/3`

`x <= 1/9`

Kết hợp với ĐKXĐ `=> 1/9 >= x > 0`

`Vậy  1/9 >= x > 0` thì `A + B <= 0`