Bài 19 : ĐK ; $ 4x \geq y^2 ; y\geq -2 $$18. \sqrt{x-1}+\sqrt{X (\sqrt{x}-1)}=x $
$19. \sqrt{4x-y^2}-\sqrt{y+2}=\sqrt{4x^2+y}$
Chuyển $ \sqrt { y + 2 } $ sang vế phải rồi bình phương lên !
Ta có : $ <=> 4x - y^2 = 4x^2 + y + y + 2 + 2 \sqrt{ ( 4x^2 + y )( y + 2 )} = 0 $
$ => ( 4x^2 - 4x + 1 ) + ( y^2 + 2y + 1 ) + 2 \sqrt{ ( 4x^2 + y )( y + 2 )} = 0 $
$ => ( 2x - 1 )^2 + ( y + 1 )^2 + 2 \sqrt{ ( 4x^2 + y )( y + 2 )} = 0 $
=> $ \left\{\begin{array}{l} ( 2x - 1 )^2 = 0\\ ( y + 1 )^2 = 0 \\ 2 \sqrt{ ( 4x^2 + y )( y + 2 )} = 0\end{array}\right. => \left\{\begin{array}{l} x = \dfrac{1}{2}\\y = -1 \end{array}\right. $