ГДЗ по алгебре 9 класс Макарычев Задание 90

 

\[\boxed{\text{90\ (90).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = \frac{1}{4}x^{2}\]

\[\textbf{а)}\ y( - 2,5) = \frac{1}{4} \cdot ( - 2,5)^{2} =\]

\[= \frac{{2,5}^{2}}{4} = 1,5625;\]

\[y( - 1,5) = \frac{1}{4} \cdot ( - 1,5)^{2} = 0,5625;\]

\[y(3,5) = \frac{1}{4} \cdot {3,5}^{2} = 3,0625.\]

\[\textbf{б)}\ y = 5:\]

\[5 = \frac{1}{4}x^{2}\ \]

\[x^{2} = 20\ \ \]

\[x = \pm \sqrt{20}\text{\ \ }\]

\[x = \pm 2\sqrt{5}.\]

\[y = 3:\]

\[3 = \frac{1}{4}x^{2}\text{\ \ }\]

\[x^{2} = 12\]

\[x = \pm \sqrt{12}\text{\ \ }\]

\[x = \pm 2\sqrt{3}.\]

\[y = 2:\]

\[2 = \frac{1}{4}x^{2}\ \]

\[x^{2} = 8\ \ \]

\[x = \pm 2\sqrt{2}.\]

\[\textbf{в)}\ Функция\ убывает\ на\ \]

\[промежутке\ ( - \infty;0\rbrack\ и\ \]

\[возрастает\]

\[на\ промежутке\ \lbrack 0;\ + \infty).\]

\[\boxed{\text{90.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \left( \sqrt{2} + \sqrt{3} \right)\left( \sqrt{2} - \sqrt{3} \right) =\]

\[= 2 - 3 = - 1\ (рациональное).\]

\[\textbf{б)}\ \left( \sqrt{2} + 2\sqrt{3} \right)\left( \sqrt{2} - \sqrt{3} \right) =\]

\[= 2 + 2\sqrt{6} - \sqrt{6} - 6 =\]

\[= \sqrt{6} - 4\ (иррациональное).\]

\[\textbf{в)}\ \frac{1^{\backslash 2 - \sqrt{3}}}{2 + \sqrt{3}} + \frac{1^{\backslash 2 + \sqrt{3}}}{2 - \sqrt{3}} =\]

\[= \frac{2 - \sqrt{3} + 2 + \sqrt{3}}{4 - 3} = \frac{4}{1} =\]

\[= 4\ (рациональное).\]

\[\textbf{г)}\ \frac{1^{\backslash\text{√}3 + \sqrt{2}}}{\sqrt{3} - \sqrt{2}} - \frac{1^{\backslash\sqrt{3} - \sqrt{2}}}{\sqrt{3} + \sqrt{2}} =\]

\[= \frac{\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}}{3 - 2} =\]

\[= 2\sqrt{2}\ (иррациональное).\]

\[\textbf{д)}\ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} =\]

\[= \frac{\left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{3} + \sqrt{2} \right)}{\left( \sqrt{3} - \sqrt{2} \right)\left( \sqrt{3} + \sqrt{2} \right)} =\]

\[= \frac{3 + 2\sqrt{6} + 2}{3 - 2} =\]

\[= 5 + 2\sqrt{6}\ (иррациональное).\]

\[\textbf{е)}\ \frac{{\sqrt{5}}^{\backslash\sqrt{5} + \sqrt{2}}}{\sqrt{5} - \sqrt{2}} + \frac{{\sqrt{5}}^{\backslash\sqrt{5} - \sqrt{2}}}{\sqrt{5} + \sqrt{2}} =\]

\[= \frac{5 + \sqrt{10} + 5 - \sqrt{10}}{\left( \sqrt{5} - \sqrt{2} \right)\left( \sqrt{5} + \sqrt{2} \right)} =\]

\[= \frac{10}{5 - 2} =\]

\[= \frac{10}{3} - рациональное.\ \]