Решебник по алгебре 8 класс Мерзляк ФГОС Задание 1127

\[\boxed{\mathbf{1127}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{\left( \sqrt{x} + 5 \right)^{2} - 20\sqrt{x}} + \sqrt{\left( \sqrt{x} - 4 \right)^{2} + 16\sqrt{x}} =\]

\[= \sqrt{\left( x + 10\sqrt{x} + 25 - 20\sqrt{x} \right)} + \sqrt{x - 8\sqrt{x} + 16 + 16\sqrt{x}} =\]

\[= \sqrt{x - 10\sqrt{x} + 25} + \sqrt{x + 8\sqrt{x} + 16} =\]

\[\sqrt{\left( \sqrt{x} - 5 \right)^{2}} + \sqrt{\left( \sqrt{x} + 4 \right)^{2}} =\]

\[= \left| \sqrt{x} - 5 \right| + \sqrt{x} + 4\]

\[x > 25:\]

\[\sqrt{x} - 5 + \sqrt{x} + 4 = 2\sqrt{x} - 1:\]

\[x = 25:\]

\[5 - \sqrt{x} + \sqrt{x} + 4 = 9;\]

\[x = 0:\]

\[5 - \sqrt{x} + \sqrt{x} + 4 = 9\]

\[2)\ \sqrt{a + 2\sqrt{a + 3} + 4} + \sqrt{a - 2\sqrt{a + 3} + 4} =\]

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

\[= \left| \sqrt{a + 3} + 1 \right| + \left| \sqrt{a + 3} - 1 \right| =\]

\[= \sqrt{a + 3} + 1 + \left| \sqrt{a + 3} - 1 \right|\]

\[- 3 \leq a \leq - 2:\]

\[\sqrt{a + 3} + 1 + 1 - \sqrt{a + 3} = 2;\]

\[a > - 2:\]

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

\[= 2\sqrt{a + 3}.\]