Задача 69. Решите уравнение \(20\,{\left( {\dfrac{{x-2}}{{x + 1}}} \right)^2}-5\,{\left( {\dfrac{{x + 2}}{{x-1}}} \right)^2} + 48\,\dfrac{{{x^2}-4}}{{{x^2}-1}} = 0\)
ОТВЕТ: 2/3; 3.
\(20\,{\left( {\dfrac{{x-2}}{{x + 1}}} \right)^2}-5\,{\left( {\dfrac{{x + 2}}{{x-1}}} \right)^2} + 48\,\dfrac{{{x^2}-4}}{{{x^2}-1}} = 0.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x + 1 \ne 0,}\\{x-1 \ne 0,}\\{{x^2}-1 \ne 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \ne -1,}\\{x \ne 1\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;x \in \left( {-\infty ;-1} \right) \cup \left( {-1;1} \right) \cup \left( {1;\infty } \right).\) \(20\,{\left( {\dfrac{{x-2}}{{x + 1}}} \right)^2}-5\,{\left( {\dfrac{{x + 2}}{{x-1}}} \right)^2} + 48\,\dfrac{{{x^2}-4}}{{{x^2}-1}} = 0\;\;\;\; \Leftrightarrow \;\;\;\;20{\left( {\dfrac{{x-2}}{{x + 1}}} \right)^2}-5{\left( {\dfrac{{x + 2}}{{x-1}}} \right)^2} + 48\dfrac{{\left( {x-2} \right)\left( {x + 2} \right)}}{{\left( {x-1} \right)\left( {x + 1} \right)}} = 0.\) Так как \(x = \pm 2\) не являются корнями исходного уравнения, то разделим обе части уравнения на \(\dfrac{{\left( {x-2} \right)\left( {x + 2} \right)}}{{\left( {x-1} \right)\left( {x + 1} \right)}}.\) Тогда уравнение примет вид: \(20\dfrac{{\left( {x-2} \right)\left( {x-1} \right)}}{{\left( {x + 2} \right)\left( {x + 1} \right)}}-5\dfrac{{\left( {x + 2} \right)\left( {x + 1} \right)}}{{\left( {x-2} \right)\left( {x-1} \right)}} + 48 = 0.\) Пусть \(\dfrac{{\left( {x-2} \right)\left( {x-1} \right)}}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = t.\) Тогда полученное уравнение примет вид: \(20t-\dfrac{5}{t} + 48 = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{20{t^2} + 48t-5 = 0,}\\{t \ne 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{t = -\dfrac{5}{2},}\\{t = \dfrac{1}{{10}}.\;}\end{array}} \right.\) Вернёмся к прежней переменной: \(\left[ {\begin{array}{*{20}{c}}{\dfrac{{\left( {x-2} \right)\left( {x-1} \right)}}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = -\dfrac{5}{2},}\\{\dfrac{{\left( {x-2} \right)\left( {x-1} \right)}}{{\left( {x + 2} \right)\left( {x + 1} \right)}} = \dfrac{1}{{10}}\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}-3x + 2}}{{{x^2} + 3x + 2}} = -\dfrac{5}{2},}\\{\dfrac{{{x^2}-3x + 2}}{{{x^2} + 3x + 2}} = \dfrac{1}{{10}}\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{2{x^2}-6x + 4 + 5{x^2} + 15x + 10}}{{2\left( {{x^2} + 3x + 2} \right)}} = 0,}\\{\dfrac{{10{x^2}-30x + 20-{x^2}-3x-2}}{{10\left( {{x^2} + 3x + 2} \right)}} = 0\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{7{x^2} + 9x + 14 = 0,}\\{3{x^2}-11x + 6 = 0\;\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{x \notin R,}\\{x = 3,\,}\end{array}}\\{x = \dfrac{2}{3}\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 3,\,}\\{x = \dfrac{2}{3}.}\end{array}} \right.\) Ответ: \(\dfrac{2}{3};\;\;\;\;3.\)