Решебник по алгебре 8 класс Мерзляк ФГОС Задание 574

 

\[\boxed{\mathbf{574\ (574).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\sqrt{10 + 8\sqrt{2 + \sqrt{9 + 4\sqrt{2}}}} =\]

\[= \sqrt{10 + 8\sqrt{2 + \sqrt{1 + 2\sqrt{8} + 8}}} =\]

\[= \sqrt{10 + 8\sqrt{2 + \sqrt{\left( 1 + \sqrt{8} \right)^{2}}}} =\]

\[= \sqrt{10 + 8\sqrt{2 + 1 + \sqrt{8}}} =\]

\[= \sqrt{10 + 8\sqrt{3 + \sqrt{8}}} =\]

\[= \sqrt{10 + 8\sqrt{1 + 2\sqrt{2} + 2}} =\]

\[= \sqrt{10 + 8\sqrt{\left( 1 + \sqrt{2} \right)^{2}}} =\]

\[= \sqrt{10 + 8 \cdot \left( 1 + \sqrt{2} \right)} =\]

\[= \sqrt{18 + 8\sqrt{2}} =\]

\[= \sqrt{\left( 2 + 8\sqrt{2} + 16 \right)} =\]

\[= \sqrt{(4 + \sqrt{2})²} = 4 + \sqrt{2}\]

\[2)\ \sqrt{22 + 6\sqrt{3 + \sqrt{13 + \sqrt{48}}}} =\]

\[= \sqrt{22 + 6\sqrt{3 + \sqrt{\left( 1 + \sqrt{12} \right)^{2}}}} =\]

\[= \sqrt{22 + 6\sqrt{3 + 1 + \sqrt{12}}} =\]

\[= \sqrt{22 + 6\sqrt{3 + 2\sqrt{3} + 1}} =\]

\[= \sqrt{22 + 6\sqrt{\left( 1 + \sqrt{3} \right)^{2}}} =\]

\[= \sqrt{22 + 6 + 6\sqrt{3}} =\]

\[= \sqrt{28 + 6\sqrt{3}} =\]

\[= \sqrt{27 + 2\sqrt{27} + 1} =\]

\[= \sqrt{\left( 1 + \sqrt{27} \right)^{2}} = 1 + \sqrt{27} =\]

\[= 1 + 3\sqrt{3}\]

\[\boxed{\mathbf{5}\mathbf{7}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{3a^{2}} = a\sqrt{3};\ \ a \geq 0\ \]

\[2)\ \sqrt{5b^{2}} = |b|\sqrt{5} = - b\sqrt{5},\]

\[b \leq 0;\]

\[3)\ \sqrt{12a^{4}} = 2a²\sqrt{3};\ \]

\[4)\ \sqrt{c^{5}} = \sqrt{c^{4} \cdot c} = c²\sqrt{c}.\]