ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 803

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

\[= \frac{10x}{3x^{2} - 2}\text{\ \ \ \ \ \ }\]

\[\ \ | \cdot \left( 3x^{2} - 2 \right),\ \ при\ x \neq \pm \sqrt{\frac{2}{3}}\]

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

\[+ \left( x\sqrt{3} - \sqrt{2} \right)^{2} = 10x\]

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

\[- 2x\sqrt{6} + 2 = 10x\]

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

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

\[D = 25 - 24 = 1\]

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

\[Ответ:x = \frac{2}{3};\ \ x = 1.\]

\[\textbf{б)}\frac{1 - y\sqrt{5}}{1 + y\sqrt{5}} + \frac{1 + y\sqrt{5}}{1 - y\sqrt{5}} =\]

\[= \frac{9y}{1 - 5y^{2}}\ \ \ \ \ \ | \cdot \left( 1 - 5y^{2} \right),\]

\[\ \ при\ y \neq \pm \sqrt{\frac{1}{5}}\]

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

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

\[+ 2y\sqrt{5} + 5y^{2} = 9y\]

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

\[D = 81 - 80 = 1\]

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

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