Решебник по алгебре 8 класс Макарычев ФГОС Задание 532

 

\[\boxed{\text{532\ (532).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\frac{9 + 6x + x^{2}}{x + 3} + \sqrt{x} =\]

\[= \frac{(3 + x)^{2}}{x + 3} + \sqrt{x} =\]

\[= \frac{(x + 3)^{2}}{x + 3} + \sqrt{x} = x + 3 + \sqrt{x}\]

\[при\ x = 0,36:\ \ \ \ \ \ \ \ \ \]

\[0,36 + 3 + \sqrt{0,36} =\]

\[= 0,36 + 3 + 0,6 = 3,96.\]

\[при\ x = 49:\ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[49 + 3 + \sqrt{49} = 49 + 3 + 7 = 59.\ \]

\[\boxed{\text{532.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 3x^{2} - 7x + 4 = 0\]

\[D = b^{2} - 4ac =\]

\[= ( - 7)^{2} - 4 \cdot 3 \cdot 4 = 49 - 48 =\]

\[= 1\]

\[x_{1,2} = \frac{- b \pm \sqrt{D}}{2a} = \frac{- ( - 7) \pm \sqrt{1}}{2 \cdot 3}\]

\[x_{1} = \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3};\ \ \ \ \ x_{2} = \frac{6}{6} = 1\]

\[Ответ:x = 1\frac{1}{3};\ \ \ x = 1.\ \]

\[\textbf{б)}\ 5x^{2} - 8x + 3 = 0\]

\[D = 64 - 4 \cdot 5 \cdot 3 = 64 - 60 = 4\]

\[x_{1,2} = \frac{8 \pm \sqrt{4}}{2 \cdot 5} = \frac{8 \pm 2}{10}\]

\[x_{1} = \frac{6}{10} = 0,6;\ \ \ \ \ \ x_{2} = \frac{10}{10} = 1\ \]

\[Ответ:x = 1;\ \ x = 0,6.\]

\[\textbf{в)}\ 3x^{2} - 13x + 14 = 0\]

\[D = 169 - 4 \cdot 3 \cdot 14 =\]

\[= 169 - 168 = 1\]

\[x_{1,2} = \frac{13 \pm \sqrt{1}}{2 \cdot 3} = \frac{13 \pm 1}{6}\]

\[x_{1} = \frac{12}{6} = 2;\ \ \]

\[\text{\ \ \ }x_{2} = \frac{14}{6} = \frac{7}{3} = 2\frac{1}{3}\ \]

\[Ответ:x = 2;\ \ x = 2\frac{1}{3}.\]

\[\textbf{г)}\ 2y^{2} - 9y + 10 = 0\]

\[D = 81 - 4 \cdot 2 \cdot 10 = 81 - 80 =\]

\[= 1\]

\[y_{1,2} = \frac{9 \pm \sqrt{1}}{2 \cdot 2} = \frac{9 \pm 1}{4}\]

\[y_{1} = \frac{8}{4} = 2;\ \ \ y_{2} = \frac{10}{4} = \frac{5}{2} = 2,5\ \]

\[Ответ:y = 2;\ \ \ y = 2,5.\]

\[\textbf{д)}\ 5y^{2} - 6y + 1 = 0\]

\[D = 36 - 4 \cdot 5 \cdot 1 = 36 - 20 =\]

\[= 16\]

\[y_{1,2} = \frac{6 \pm \sqrt{16}}{2 \cdot 5} = \frac{6 \pm 4}{10}\]

\[y_{1} = \frac{2}{10} = 0,2;\ \ \ \ \ y_{2} = \frac{10}{10} = 1\ \]

\[Ответ:y = 0,2;\ \ y = 1.\]

\[\textbf{е)}\ 4x^{2} + x - 33 = 0\]

\[D = 1 - 4 \cdot 4 \cdot ( - 33) = 1 + 528 =\]

\[= 529\]

\[x_{1,2} = \frac{- 1 \pm \sqrt{529}}{2 \cdot 4} = \frac{- 1 \pm 23}{8}\]

\[x_{1} = \frac{22}{8} = \frac{11}{4} = 2\frac{3}{4} = 2,75;\ \ \ \]

\[x_{2} = - \frac{24}{8} = - 3\ \]

\[Ответ:x = 2,75;\ \ \ x = - 3.\]

\[\textbf{ж)}\ y^{2} - 10y - 24 = 0\]

\[D = 100 - 4 \cdot ( - 24) =\]

\[= 100 + 96 = 196\]

\[y_{1,2} = \frac{10 \pm \sqrt{196}}{2} = \frac{10 \pm 14}{2}\]

\[y_{1} = \frac{24}{2} = 12;\ \ \ \ y_{2} = - \frac{4}{2} = - 2\ \]

\[Ответ:y = 12;\ \ y = - 2.\]

\[\textbf{з)}\ p^{2} + p - 90 = 0\]

\[D = 1 - 4 \cdot ( - 90) = 1 + 360 =\]

\[= 361\]

\[p_{1,2} = \frac{- 1 \pm \sqrt{361}}{2} = \frac{- 1 \pm 19}{2}\]

\[p_{1} = \frac{18}{2} = 9;\ \ \ \ \ p_{2} = - \frac{20}{2} =\]

\[= - 10\ \ \ \]

\[Ответ:p = 9;\ \ p = - 10.\]