Đáp án+Giải thích các bước giải:

ĐKXĐ `x >= 2 ; 2 >= y`

`{( 2x + 2y + \sqrt{( 2x - 1 )( 2 - y )} = 5 ( 1 ) ),( \sqrt{x^2 + x - 6} + 3\sqrt{x - 1} = \sqrt{y^2 - x^2 - 3y + 19} ( 2 ) ):}`

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

Đặt `\sqrt{2x - 1} = a ; \sqrt{2 - y} = b  ( a ; b >= 0 )`

pt `( 1 ) <=> a^2 + ab - 2b^2 = 0`

`( a - b )( a + 2b ) = 0`

`a - b = 0  ( 3 )  hoặc   a + 2b = 0  ( 4 )`

`( 4 ) <=> a = b = 0 => x = 1/2 ; y = 2` ( thử lại KTM )

`( 3 ) <=> a = b <=> 2x - 1 = 2 - y => y = 3 - 2x ( 5 )`

`( 5 ) ; ( 2 ) => \sqrt{x^2 + x - 6} + 3\sqrt{x - 1} = \sqrt{9 - 12x + 4x^2 - x^2 - 9 + 6x + 19}`

`\sqrt{x^2 + x - 6} + 3\sqrt{x - 1} = \sqrt{3x^2 - 6x + 19}`

 `x^2 + x - 6 + 6\sqrt{( x - 1 )( x^2 + x - 6 )} + 9x - 9 = 3x^2 - 6x + 19`

`6\sqrt{( x - 1 )( x -2 )( x + 3 )} = 2x^2 - 16x + 34`

`3\sqrt{( x^2 + 2x - 3 )( x - 2 )} = x^2 - 8x + 17`

`3\sqrt{( x^2 + 2x - 3 )( x - 2 )} = x^2 + 2x - 3 - 10( x - 2 )`

Đặt `\sqrt{x^2 + 2x - 3} = a ; \sqrt{x - 2} = b  ( a ; b >= 0 )`

pt `<=> a^2 - 3ab - 10b^2 = 0`

`( a + 2b )( a - 5b ) = 0`

`a + 2b = 0  ( 6 )  hoặc   a - 5b = 0  ( 7 )`

`( 6 ) <=> a = b = 0 => x^2 + 2x - 3 = x - 2 => x^2 + x + 1/4 = 5/4`

`( x + 1/2 )^2 = 5/4`

`x = ( -1 +- \sqrt{5} )/2` ( KTM )

`( 7 ) <=> a = 5b`

`<=> x^2 + 2x - 3 = 25x - 50`

`<=> x^2 - 22x + 47 = 0`

`<=> x^2 - 2 . 11x + 121 = 74`

`<=> ( x - 11 )^2 = 74`

`<=> x = 11 +- \sqrt{74}`

Thử lại TM

`x = 11 + \sqrt{74} => y = -19 - 2\sqrt{74}` ( TM )

`x = 11 - \sqrt{74} => y = -19 + 2\sqrt{74}` ( TM )

`Vậy  ( x ; y ) in { ( 11 + \sqrt{74} ; -19 - 2\sqrt{74} ) ; ( 11 - \sqrt{74} ; -19 + 2\sqrt{74} ) }`