\[\sqrt {x - 3} = 2 - x\]
\[\sqrt {x - 1} + \sqrt {x + 3} = 2\]
\[\sqrt {5 - x} + \sqrt {5x - 5} = 2\]
\[\sqrt {{x^2} - 2x} + \sqrt {x - {x^2}} = \sqrt x \]
\[\sqrt {2x + 3} - \sqrt {4 - x} = 2\]
\[\sqrt[6]{{1 - x}} - 5\sqrt[3]{{1 - x}} = - 6\]
\[{\text{Решите уравнение }}x + \sqrt {{x^4} - 2x - 19} = 1\]
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\[\sqrt {f\left( x \right)} > g\left( x \right) \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} f\left( x \right) > {g^2}\left( x \right) \hfill \\ g\left( x \right) \geqslant 0 \hfill \\ \end{array} \right. \hfill \\ \left\{ \begin{array}{l} f\left( x \right) \geqslant 0 \hfill \\ g\left( x \right) < 0 \hfill \\ \end{array} \right. \hfill \\ \end{array} \right.\]
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\[\sqrt {f\left( x \right)} < g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l} f\left( x \right) < {g^2}\left( x \right) \hfill \\ g\left( x \right) > 0 \hfill \\ f\left( x \right) \geqslant 0 \hfill \\ \end{array} \right.\]
\[\sqrt {x + 1} < x - 1\]
\[\sqrt {25 - {x^2}} \geqslant 4\]
\[\sqrt {{x^2} - 3x} \leqslant \sqrt {3x + 7} \]