\[\boxed{\text{427\ (427).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} - 7 = x^{2} - \left( \sqrt{7} \right)^{2} =\]
\[= \left( x - \sqrt{7} \right)\left( x + \sqrt{7} \right)\]
\[\textbf{б)}\ 5 - c^{2} = \left( \sqrt{5} \right)^{2} - c^{2} =\]
\[= \left( \sqrt{5} - c \right)\left( \sqrt{5} + c \right)\]
\[\textbf{в)}\ 4a^{2} - 3 = (2a)^{2} - \left( \sqrt{3} \right)^{2} =\]
\(= (2a - \sqrt{3})(2a + \sqrt{3})\)
\[\textbf{г)}\ 11 - 16b^{2} =\]
\[= \left( \sqrt{11} \right)^{2} - (4b)^{2} =\]
\[= \left( \sqrt{11} - 4b \right)\left( \sqrt{11} + 4b \right)\ \]
\[\textbf{д)}\ y \geq 0:\ \ \]
\[y - 3 = \left( \sqrt{y} \right)^{2} - \left( \sqrt{3} \right)^{2} =\]
\[= \left( \sqrt{y} - \sqrt{3} \right)\left( \sqrt{y} + \sqrt{3} \right)\]
\[\textbf{е)}\ x > 0;\ y > 0:\ \ \]
\[x - y = \left( \sqrt{x} \right)^{2} - \left( \sqrt{y} \right)^{2} =\]
\[= \left( \sqrt{x} - \sqrt{y} \right)\left( \sqrt{x} + \sqrt{y} \right)\text{\ \ }\]