Задача 44. Решите уравнение \(\dfrac{{\sqrt {x-1} + \sqrt {x + 1} + 2}}{{\sqrt {x-1} + \sqrt {x + 1} }} = \sqrt {{x^2}-1} .\)
ОТВЕТ: 2.
\(\dfrac{{\sqrt {x-1} + \sqrt {x + 1} + 2}}{{\sqrt {x-1} + \sqrt {x + 1} }} = \sqrt {{x^2}-1} .\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{x-1 \ge 0,}\\{x + 1 \ge 0,}\\{{x^2}-1 \ge 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left[ {1;\infty } \right).\) \(\dfrac{{\sqrt {x-1} + \sqrt {x + 1} + 2}}{{\sqrt {x-1} + \sqrt {x + 1} }} = \sqrt {{x^2}-1} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\sqrt {x-1} + \sqrt {x + 1} }}{{\sqrt {x-1} + \sqrt {x + 1} }} + \dfrac{2}{{\sqrt {x-1} + \sqrt {x + 1} }} = \sqrt {{x^2}-1} \,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,1 + \dfrac{{2\left( {\sqrt {x-1} -\sqrt {x + 1} } \right)}}{{\left( {\sqrt {x-1} + \sqrt {x + 1} } \right)\left( {\sqrt {x-1} -\sqrt {x + 1} } \right)}} = \sqrt {{x^2}-1} \,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,1 + \dfrac{{2\left( {\sqrt {x-1} -\sqrt {x + 1} } \right)}}{{x-1-x-1}} = \sqrt {{x^2}-1} \,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,1-\sqrt {x-1} + \sqrt {x + 1} = \sqrt {{x^2}-1} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {x + 1} -\sqrt {x-1} = \sqrt {{x^2}-1} -1\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sqrt {{x^2}-1} -1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x + 1-2\sqrt {{x^2}-1} + x-1 = {x^2}-1-2\sqrt {{x^2}-1} + 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\sqrt {{x^2}-1} \ge 1,}\\{{x^2}-2x = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left( {-\infty ;-\sqrt 2 } \right] \cup \left[ {\sqrt 2 ;\infty } \right),}\\{\left[ {\begin{array}{*{20}{c}}{x = 0,}\\{x = 2\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 2.\) Ответ: 2.