Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1569

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\[1)\ \left\{ \begin{matrix} \log_{y}x + \log_{x}y = \frac{5}{2} \\ x + y = a + a^{2}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\log_{y}x + \frac{\log_{y}y}{\log_{y}x}\text{-}\frac{5}{2} = 0\]

\[\log_{y}x + \frac{1}{\log_{y}x} - \frac{5}{2} = 0\ \ \ \ \ | \bullet 2\log_{y}x\]

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

\[u = \log_{y}x:\]

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

\[D = 25 - 16 = 9\]

\[u_{1} = \frac{5 - 3}{2 \bullet 2} = \frac{1}{2};\ u_{2} = \frac{5 + 3}{2 \bullet 2} = 2.\]

\[1)\ \log_{y}x = \frac{1}{2}\]

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

\[x = \sqrt{y}.\]

\[2)\ \log_{y}x = 2\]

\[\log_{y}x = \log_{y}y^{2}\]

\[x = y^{2}.\]

\[Первая\ система:\]

\[\left\{ \begin{matrix} x = \sqrt{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x + y = a + a^{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} y = x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x + y - \left( a + a^{2} \right) = 0 \\ \end{matrix} \right.\ \]

\[x^{2} + x - \left( a + a^{2} \right) = 0\]

\[D = 1^{2} + 4\left( a + a^{2} \right) =\]

\[= 1 + 4a + 4a^{2} = (2a + 1)^{2}\]

\[x_{1} = \frac{- 1 - (2a + 1)}{2} =\]

\[= \frac{- 2 - 2a}{2} = - a - 1;\]

\[x_{2} = \frac{- 1 + 2a + 1}{2} = \frac{2a}{2} = a;\]

\[y_{1} = ( - a - 1)^{2} = (a + 1)^{2};\]

\[y_{2} = a^{2}.\]

\[Вторая\ система:\]

\[\left\{ \begin{matrix} x = y^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x + y = a + a^{2} \\ \end{matrix} \right.\ \]

\[y^{2} + y - \left( a + a^{2} \right) = 0\]

\[D = 1^{2} + 4\left( a + a^{2} \right) =\]

\[= 1 + 4a + 4a^{2} = (2a + 1)^{2}\]

\[y_{1} = \frac{- 1 - (2a + 1)}{2} =\]

\[= \frac{- 2 - 2a}{2} = - a - 1;\]

\[y_{2} = \frac{- 1 + 2a + 1}{2} = \frac{2a}{2} = a;\]

\[x_{1} = ( - a - 1)^{2} = (a + 1)^{2};\]

\[x_{2} = a^{2}.\]

\[Имеет\ смысл\ при:\]

\[a > 0\]

\[- a - 1 > 0\]

\[\ a < - 1.\]

\[a \neq 1\]

\[- a - 1 \neq 1\]

\[a \neq - 2.\]

\[2)\ \left\{ \begin{matrix} x^{2} + y^{2} = a^{2}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ \log_{b}x + \log_{b}y = 2 \\ \end{matrix} \right.\ \]

\[Второе\ уравнение:\]

\[\log_{b}\text{xy} = 2\]

\[\log_{b}\text{xy} = \log_{b}b^{2}\]

\[xy = b^{2}\]

\[x = \frac{b^{2}}{y}.\]

\[Подставим\ в\ первое\ уравнение:\]

\[\left( \frac{b^{2}}{y} \right)^{2} + y^{2} = a^{2}\]

\[\frac{b^{4}}{y^{2}} + y^{2} = a^{2}\ \ \ \ \ | \bullet y^{2}\]

\[b^{4} + y^{4} = a^{2}y^{2}\]

\[y^{4} - a^{2}y^{2} + b^{4} = 0\]

\[z = y^{2}:\]

\[z^{2} - a^{2}z + b^{4} = 0\]

\[D = \left( a^{2} \right)^{2} - 4b^{4} = a^{4} - 4b^{4}\]

\[z = \frac{a^{2} \pm \sqrt{a^{4} - 4b^{4}}}{2};\]

\[y = \pm \sqrt{\frac{a^{2} \pm \sqrt{a^{4} - 4b^{4}}}{2}};\]

\[x = b^{2}\ :\left( \pm \sqrt{\frac{a^{2} \pm \sqrt{a^{4} - 4b^{4}}}{2}} \right) =\]

\[= \pm \sqrt{\frac{2b^{4}}{a^{2} \pm \sqrt{a^{4} - 4b^{4}}}}.\ \]

\[Имеет\ смысл\ при:\]

\[b \neq 1;\ \ b \neq 0;\]

\[x > 0;\ \ y > 0.\]

\[a^{2} \pm \sqrt{a^{4} - 4b^{4}} > 0\]

\[a^{4} \pm \left( a^{4} - 4b^{4} \right) > 0\]

\[4b^{4} > 0\]

\[b \neq 0.\]

\[2a^{4} - 4b^{4} > 0\]

\[2\left( a^{2} - \sqrt{2}b^{2} \right)\left( a^{2} + \sqrt{2}b^{2} \right) > 0\]

\[a^{2} - \sqrt{2}b^{2} > 0.\]

\[если\ a^{2} - \sqrt{2}b^{2} > 0,\ b \neq 0,\ b \neq 1.\]