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+) $a ≥ 0 ⇒ | a | = a$

+) $a < 0 ⇒ | a | = - a$

$1/ \sqrt[]{3x-2} = \sqrt[]{28-10\sqrt[]{3}} - \sqrt[]{19-8\sqrt[]{3}}$ $( x ≥ \frac{2}{3} )$

⇔ $\sqrt[]{3x-2} = \sqrt[]{25-10\sqrt[]{3}+3} - \sqrt[]{16-8\sqrt[]{3}+3}$

⇔ $\sqrt[]{3x-2} = \sqrt[]{(5-\sqrt[]{3})^{2}} - \sqrt[]{(4-\sqrt[]{3})^{2}}$

⇔ $\sqrt[]{3x-2} = | 5 - \sqrt[]{3} | - | 4 - \sqrt[]{3} |$

⇔ $\sqrt[]{3x-2} = 5 - \sqrt[]{3} - 4 + \sqrt[]{3}$

⇔ $\sqrt[]{3x-2} = 1$

⇔ $3x - 2 = 1$

⇔ $3x = 3$

⇔ $x = 1$ ( TM )
$2.\sqrt[]{x^{2}-7x+16} = \sqrt[]{11-6\sqrt[]{2}} + \sqrt[]{2\sqrt[]{2}+3}$

⇔ $\sqrt[]{x^{2}-7x+16} = \sqrt[]{9-6\sqrt[]{2}+2} + \sqrt[]{2+2\sqrt[]{2}+1}$

⇔ $\sqrt[]{x^{2}-7x+16} = \sqrt[]{(3-\sqrt[]{2})^{2}} + \sqrt[]{(\sqrt[]{2}+1)^{2}}$

⇔ $\sqrt[]{x^{2}-7x+16} = | 3 - \sqrt[]{2} | + | \sqrt[]{2} + 1 |$

⇔ $\sqrt[]{x^{2}-7x+16} = 3 - \sqrt[]{2} + \sqrt[]{2} + 1$

⇔ $\sqrt[]{x^{2}-7x+16} = 4$

⇔ $x^{2} - 7x + 16 = 16$

⇔ $x^{2} - 7x = 0$

⇔ $x( x - 7 ) = 0$

⇔ \(\left[ \begin{array}{l}x=0\\x=7\end{array} \right.\) 

$3. \sqrt[]{x^{2}-x-3} = \sqrt[]{2\sqrt[]{5}+6} - \sqrt[]{(\sqrt[]{5}-2)^{2}}$ $( x^{2} - x - 3 ≥ 0 )$

⇔ $\sqrt[]{x^{2}-x-3} = \sqrt[]{5+2\sqrt[]{5}+1} - | \sqrt[]{5} - 2 |$

⇔ $\sqrt[]{x^{2}-x-3} = \sqrt[]{(\sqrt[]{5}+1)^{2}} - ( \sqrt[]{5} - 2 )$

⇔ $\sqrt[]{x^{2}-x-3} = | \sqrt[]{5} + 1 | - \sqrt[]{5} + 2$

⇔ $\sqrt[]{x^{2}-x-3} = \sqrt[]{5} + 1 - \sqrt[]{5} + 2$

⇔ $\sqrt[]{x^{2}-x-3} = 3$

⇔ $x^{2} - x - 3 = 9$

⇔ $x^{2} - x - 12 = 0$

⇔ $( x - 4 )( x + 3 ) = 0$

⇔ \(\left[ \begin{array}{l}x=4\\x=-3\end{array} \right.\) ( TM )

$4. \sqrt[]{x^{2}-6x+9} = \sqrt[]{16-6\sqrt[]{7}} + \sqrt[]{2\sqrt[]{7}+8}$

⇔ $\sqrt[]{(x-3)^{2}} = \sqrt[]{9-6\sqrt[]{7}+7} + \sqrt[]{7+2\sqrt[]{7}+1}$

⇔ $| x - 3 | = \sqrt[]{(3-\sqrt[]{7})^{2}} + \sqrt[]{(\sqrt[]{7}+1)^{2}}$

⇔ $| x - 3 | = | 3 - \sqrt[]{7} | + | \sqrt[]{7} + 1 |$

⇔ $| x - 3 | = 3 - \sqrt[]{7} + \sqrt[]{7} + 1$

⇔ $| x - 3 | = 4$

⇔ $x - 3 = ±4$

⇔ \(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\)