Решебник по алгебре 8 класс Макарычев ФГОС Задание 695

 

\[\boxed{\text{695\ (695).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[= 10x\]

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

\[= 10x\]

\[6x^{2} - 10x + 4 = 0\ \ \ \ \ \ \ |\ :2\]

\[3x^{2} - 5x + 2 = 0\]

\[D = 25 - 24 = 1\]

\[x_{1,2} = \frac{5 \pm 1}{6} = 1;\frac{2}{3}\]

\[Ответ:x = \frac{2}{3};\ \ x = 1.\]

\[\left( 1 - y\sqrt{5} \right)^{2} + \left( 1 + y\sqrt{5} \right)^{2} = 9y\]

\[1 - 2y\sqrt{5} + 5y^{2} + 1 + 2y\sqrt{5} + 5y^{2} =\]

\[= 9y\]

\[10y^{2} - 9y + 2 = 0\]

\[D = 81 - 80 = 1\]

\[y_{1,2} = \frac{9 \pm 1}{20} = \frac{1}{2};\frac{2}{5}\]

\[Ответ:y = 0,4;\ \ y = 0,5\text{.\ }\]

\[\boxed{\text{695.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ 16m^{2} - 25n^{4} =\]

\[= \left( 4m - 5n^{2} \right)\left( 4m + 5n^{2} \right)\]

\[\textbf{б)}\ 0,09a^{4} - 9b^{2} =\]

\[= (0,3a^{2} - 3b)(0,3a^{2} + 3b)\]

\[\textbf{в)}\ (2a + 3)^{2} - 4 =\]

\[= (2a + 3 - 2)(2a + 3 + 2) =\]

\[= (2a + 1)(2a + 5)\]

\[\textbf{г)}\ 36 - (1 - 5x)^{2} =\]

\[= \left( 6 - (1 - 5x) \right)(6 + 1 - 5x) =\]

\[= (5 + 5x)(7 - 5x)\]