ГДЗ по алгебре 8 класс Макарычев Задание 452

\[\boxed{\text{452\ (452).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b \geq 1:\]

\[\ \sqrt{\frac{b + 1}{2} - {\sqrt{b}}^{\backslash 2}} - \sqrt{\frac{b + 1}{2} + {\sqrt{b}}^{\backslash 2}} =\]

\[= \sqrt{\frac{b + 1 - 2\sqrt{b}}{2}} - \sqrt{\frac{b + 1 + 2\sqrt{b}}{2}} =\]

\[= \sqrt{\frac{\left( \sqrt{b} - 1 \right)^{2}}{2}} - \sqrt{\frac{\left( \sqrt{b} + 1 \right)^{2}}{2}} =\]

\[= \frac{\sqrt{b} - 1}{\sqrt{2}} - \frac{\sqrt{b} + 1}{\sqrt{2}} =\]

\[= \frac{\sqrt{b} - 1 - \sqrt{b} - 1}{\sqrt{2}} =\]

\[= - \frac{2}{\sqrt{2}} = \frac{- 2\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = - \frac{2\sqrt{2}}{2} =\]

\[= - \sqrt{2}\]

\[\textbf{б)}\ c \geq 4\ \]

\[\sqrt{\frac{c + 4}{4} + {\sqrt{c}}^{\backslash 4}} - \sqrt{\frac{c + 4}{4} - {\sqrt{c}}^{\backslash 4}} =\]

\[= \sqrt{\frac{c + 4 + 4\sqrt{c}}{4}} - \sqrt{\frac{c + 4 - 4\sqrt{c}}{4}} =\]

\[= \sqrt{\frac{\left( \sqrt{c} + 2 \right)^{2}}{4}} - \sqrt{\frac{\left( \sqrt{c} - 2 \right)^{2}}{4}} =\]

\[= \frac{\sqrt{c} + 2}{2} - \frac{\sqrt{c} - 2}{2} =\]

\[= \frac{\sqrt{c} + 2 - \sqrt{c} + 2}{2} = \frac{4}{2} = 2\]