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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 596
Задание 596
\[\boxed{\text{596.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ (3x + 1)^{2} = 3x + 1\] \[9x^{2} + 6x + 1 = 3x + 1\] \[9x^{2} + 3x = 0\] \[3x(3x + 1) = 0\] \[3x = 0\ \ \ \ \ \ 3x = - 1\] \[\ \ x = 0\ \ \ \ \ \ \ x = - \frac{1}{3}\] \[Ответ:при\ x = - \frac{1}{3};x = 0.\] |
\[\textbf{г)}\ (3x + 4)^{2} = 4(x + 3)\] \[9x^{2} + 24x + 16 = 4x + 12\] \[9x^{2} + 20x + 4 = 0\] \[D = 400 - 144 = 256\] \[x_{1,2} = \frac{- 20 \pm \sqrt{256}}{18} = \frac{- 20 \pm 16}{18}\] \[x_{1} = - \frac{2}{9};\ \ x_{2} = - 2.\] \[Ответ:при\ x = - 2;\ \ x = - \frac{2}{9}.\] |
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\[\textbf{б)}\ (3x + 1)^{2} = 3(x + 1)\] \[9x^{2} + 6x + 1 = 3x + 3\] \[9x^{2} + 3x - 2 = 0\] \[D = 9 + 72 = 81\] \[x_{1,2} = \frac{- 3 \pm \sqrt{81}}{18} = \frac{- 3 \pm 9}{18}\] \[x_{1} = - \frac{2}{3};\ \ \ x_{2} = \frac{1}{3}.\] \[Ответ:при\ x = - \frac{2}{3};\ \ x = \frac{1}{3}.\] |
\[\textbf{д)}\ 4(x + 3)^{2} = (2x + 6)^{2}\] \[4\left( x^{2} + 6x + 9 \right) = 4x^{2} + 24x + 36\] \[4x^{2} + 24x + 36 = 4x^{2} + 24x + 36\] \[x - любое\ число.\] \[Ответ:при\ x - любое\ число.\] |
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\[\textbf{в)}\ (3x + 1)^{2} = (2x - 5)^{2}\] \[9x^{2} + 6x + 1 = 4x^{2} - 20x + 25\] \[5x^{2} + 26x - 24 = 0\] \[D = 676 + 480 = 1156\] \[x_{1,2} = \frac{- 26 \pm \sqrt{1156}}{10} = \frac{- 26 \pm 34}{10}\] \[x_{1} = 0,8;\ \ x_{2} = - 6.\] \[Ответ:при\ x = - 6;\ \ x = 0,8.\] |
\[\textbf{е)}\ (6x + 3)^{2} = (x - 4)^{2}\] \[36x^{2} + 36x + 9 = x^{2} - 8x + 16\] \[35x^{2} + 44x - 7 = 0\] \[D = 1936 + 980 = 2916\] \[x_{1,2} = \frac{- 44 \pm \sqrt{2916}}{70} = \frac{- 44 \pm 54}{70}\] \[x_{1} = \frac{1}{7};\ \ x_{2} = - 1,4.\ \text{\ \ \ }\] \[Ответ:при\ x = - 1,4;\ \ x = \frac{1}{7}.\] |
\[\boxed{\text{596.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[І\ способ.\ \]
\[\frac{c}{a} = \frac{13x}{12x},\ \ b = 15\ см\]
\[По\ теореме\ Пифагора:\]
\[15^{2} + (12x)^{2} = (13x)^{2}\]
\[225 + 144x^{2} = 169x^{2}\]
\[25x^{2} = 225\]
\[x^{2} = 9\]
\[x = \sqrt{9} = \pm 3\]
\[x = 3,\ \ так\ как - 3 \notin N\]
\[a = 12 \cdot 3 = 36\ см\]
\[b = 13 \cdot 3 = 39\ см\]
\[P = a + b + c = 15 + 36 + 39 =\]
\[= 90\ (см).\]
\[Пусть\ 13\text{x\ }см - гипотенуза,\ а\ \]
\[12x\ см - катет.\]
\[Составим\ уравнение,\]
\[используя\ теорему\ Пифагора:\]
\[(12x)^{2} + 15^{2} = (13x)^{2}\]
\[144x^{2} + 225 = 169x^{2}\]
\[169x^{2} - 144x^{2} = 225\]
\[25x^{2} = 225\]
\[x^{2} = 9\]
\[x = 3\ (см).\]
\[13x = 3 \cdot 13 = 39\ (см) -\]
\[гипотенуза.\]
\[12x = 12 \cdot 3 = 36\ (см) - катет.\]
\[P = 39 + 36 + 15 = 90\ см.\]
\[Ответ:90\ см.\ \ \]
