Решебник по алгебре 8 класс Мерзляк ФГОС Задание 529

\[\boxed{\mathbf{529\ (529).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 4\sqrt{a} + 3\sqrt{a} - 5\sqrt{a} =\]

\[= \sqrt{a} \cdot (4 + 3 - 5) = 2\sqrt{a};\]

\[2)\ 6\sqrt{b} + 2\sqrt{b} - 8\sqrt{b} =\]

\[= \sqrt{b} \cdot (6 + 2 - 8) = 0;\]

\[3)\ 5\sqrt{c} + 3\sqrt{d} - \sqrt{c} + 3\sqrt{d} =\]

\[= 4\sqrt{c} + 6\sqrt{d};\]

\[4)\ \sqrt{5} + 7\sqrt{5} - 4\sqrt{5} = 4\sqrt{5}.\]

\[\boxed{\mathbf{5}\mathbf{2}\mathbf{9}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{\sqrt{48}}{\sqrt{3}} = \sqrt{16} = 4\]

\[2)\ \frac{\sqrt{150}}{\sqrt{6}} = \sqrt{25} = 5\]

\[3)\ \frac{\sqrt{6,3}}{\sqrt{0,7}} = \sqrt{9} = 3\]

\[4)\ \frac{\sqrt{98}}{\sqrt{242}} = \sqrt{\frac{98}{242}} = \sqrt{\frac{49}{121}} = \frac{7}{11}\]

\[5)\ \frac{\sqrt{6} \cdot \sqrt{2}}{\sqrt{3}} = \frac{\sqrt{12}}{\sqrt{3}} = \sqrt{4} = 2\]