ГДЗ по алгебре 9 класс Макарычев Задание 20

 

\[\boxed{\text{20\ (20).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = \frac{x^{2}}{x^{2} + 1} = \frac{x^{2} + 1 - 1}{x^{2} + 1} =\]

\[= \frac{x^{2} + 1}{x^{2} + 1} - \frac{1}{x^{2} + 1} = 1 - \frac{1}{x^{2} + 1}.\]

\[x^{2} + 1 \geq 1\ при\ любом\ \]

\[значении\ x \Longrightarrow D(y) = R.\]

\[Так\ как\ \ \ x^{2} + 1 \geq 1:\]

\[0 < \frac{1}{x^{2} + 1} \leq 1\ \ \ \ \ | \cdot ( - 1)\]

\[0 > - \frac{1}{x^{2} + 1} \geq - 1\ \ \ \ \ \ | + 1\]

\[- 1 + 1 \leq - \frac{1}{x^{2} + 1} + 1 < 0 + 1\]

\[0 \leq 1 - \frac{1}{x^{2} + 1} < 1.\]

\[E(y) = \lbrack 0;1).\]

\[\boxed{\text{20.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ a = 1;\ \ b = 0,64:\]

\[\sqrt{a} - \sqrt{b} = \sqrt{1} - \sqrt{0,64} =\]

\[= 1 - 0,8 = 0,2.\]

\[\textbf{б)}\ a = 1;\ \ b = 0,64:\]

\[\sqrt{a - b} = \sqrt{1 - 0,64} = \sqrt{0,36} =\]

\[= 0,6.\]

\[\textbf{в)}\ a = 0,12;\ \ b = 0,01:\]

\[2\sqrt{a + 4b} = 2\sqrt{0,12 + 4 \cdot 0,01} =\]

\[= 2\sqrt{0,16} = 2 \cdot 0,4 = 0,8.\]

\[\textbf{г)}\ a = 0,6;\ \ b = 0,8:\]

\[\sqrt{3a - b} = \sqrt{3 \cdot 0,6 - 0,8} =\]

\[= \sqrt{1,8 - 0,8} = \sqrt{1} = 1.\]

\[\textbf{д)}\ a = 0,7;\ \ b = 0,09:\]

\[\sqrt{a + \sqrt{b}} = \sqrt{0,7 + \sqrt{0,09}} =\]

\[= \sqrt{0,7 + 0,3} = \sqrt{1} = 1.\]

\[\textbf{е)}\ a = 4,8;\ \ b = 0,64:\]

\[- \sqrt{a - \sqrt{b}} = - \sqrt{4,8 - \sqrt{0,64}} =\]

\[= - \sqrt{4,8 - 0,8} = - \sqrt{4} = - 2.\]