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МатематикаРешебник по алгебре 8 класс Мерзляк ФГОС Задание 602
Задание 602
\[\boxed{\mathbf{602\ (602).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[f = \left\{ \begin{matrix} x^{2},\ \ x \leq 1 \\ \sqrt{x},\ \ x > 1 \\ \end{matrix} \right.\ \]
\[1)f( - 2) = 4;\ \ \ \ \ \]
\[f(0) = 0;\ \ \ \ \ \]
\[f(1) = 1;\ \ \ \ \ \]
\[f(4) = 2\]
\[2)\ f = x²\]
| \[x\] | \[0\] | \[1\] | \[- 1\] | \[2\] | \[- 2\] |
|---|---|---|---|---|---|
| \[f\] | \[0\] | \[1\] | \[1\] | \[4\] | \[4\] |
\[f = \sqrt{x}\]
| \[x\] | \[1\] | \[4\] | \[9\] | \[0\] |
|---|---|---|---|---|
| \[f\] | \[1\] | \[2\] | \[3\] | \[0\] |
\[\boxed{\mathbf{60}\mathbf{2}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\sqrt{10 + 8\sqrt{2 + \sqrt{9 + 4\sqrt{2}}}} =\]
\[= \sqrt{10 + 8\sqrt{2 + \sqrt{1 + 2\sqrt{8} + 8}}} =\]
\[= \sqrt{10 + 8\sqrt{2 + \sqrt{\left( 1 + \sqrt{8} \right)^{2}}}} =\]
\[= \sqrt{10 + 8\sqrt{2 + 1 + \sqrt{8}}} =\]
\[= \sqrt{10 + 8\sqrt{3 + \sqrt{8}}} =\]
\[= \sqrt{10 + 8\sqrt{1 + 2\sqrt{2} + 2}} =\]
\[= \sqrt{10 + 8\sqrt{\left( 1 + \sqrt{2} \right)^{2}}} =\]
\[= \sqrt{10 + 8 \cdot \left( 1 + \sqrt{2} \right)} =\]
\[= \sqrt{18 + 8\sqrt{2}} =\]
\[= \sqrt{\left( 2 + 8\sqrt{2} + 16 \right)} =\]
\[= \sqrt{(4 + \sqrt{2})²} = 4 + \sqrt{2}\]
\[2)\ \sqrt{22 + 6\sqrt{3 + \sqrt{13 + \sqrt{48}}}} =\]
\[= \sqrt{22 + 6\sqrt{3 + \sqrt{\left( 1 + \sqrt{12} \right)^{2}}}} =\]
\[= \sqrt{22 + 6\sqrt{3 + 1 + \sqrt{12}}} =\]
\[= \sqrt{22 + 6\sqrt{3 + 2\sqrt{3} + 1}} =\]
\[= \sqrt{22 + 6\sqrt{\left( 1 + \sqrt{3} \right)^{2}}} =\]
\[= \sqrt{22 + 6 + 6\sqrt{3}} =\]
\[= \sqrt{28 + 6\sqrt{3}} =\]
\[= \sqrt{27 + 2\sqrt{27} + 1} =\]
\[= \sqrt{\left( 1 + \sqrt{27} \right)^{2}} = 1 + \sqrt{27} =\]
\[= 1 + 3\sqrt{3}\]
