Простейшие иррациональные уравнения. Задача 27

ОТВЕТ: 0;  7.

\(\sqrt {{x^2}-4x + 1}  = \sqrt {3x + 1} .\)

Уравнение вида:  \(\sqrt {f\left( x \right)}  = \sqrt {g\left( x \right)} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) \ge 0,\,\,\,\,\,\,}\\{f\left( x \right) = g\left( x \right)}\end{array}} \right.\)  или  \(\left\{ {\begin{array}{*{20}{c}}{g\left( x \right) \ge 0,\,\,\,\,\,\,\,\,\,}\\{f\left( x \right) = g\left( x \right).}\end{array}} \right.\)

\(\sqrt {{x^2}-4x + 1}  = \sqrt {3x + 1} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3x + 1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-4x + 1 = 3x + 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ge  — \dfrac{1}{3},\,\,\,\,\,\,}\\{{x^2}-7x = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ge -\dfrac{1}{3},}\\{\left[ {\begin{array}{*{20}{c}}{x = 0,}\\{x = 7\,}\end{array}\,} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,}\\{x = 7.}\end{array}} \right.\)

Ответ:  0; 7.