Задача 2С. Решите уравнение \(\sqrt {{x^2} — 3x + 1} = \left| {2x — 1} \right| — x.\)
ОТВЕТ: 0.
\(\sqrt {{x^2}-3x + 1} = \left| {2x-1} \right|-x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{2x-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {{x^2}-3x + 1} = 2x-1-x,}\end{array}\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{2x-1 < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {{x^2}-3x + 1} = -2x + 1-x.}\end{array}} \right.}\end{array}} \right.\) Рассмотрим первую систему, полученной совокупности: \(\left\{ {\begin{array}{*{20}{c}}{2x-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {{x^2}-3x + 1} = x-1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2x-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3x + 1 = {x^2}-2x + 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ge 0,5,}\\{x \ge 1,\,\,\,\,\,\,}\\{x = 0\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\emptyset .} \right.\) Рассмотрим вторую систему: \(\left\{ {\begin{array}{*{20}{c}}{2x-1 < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt {{x^2}-3x + 1} = 1-3x}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2x-1 < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{1-3x \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-3x + 1 = 1-6x + 9{x^2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x < \dfrac{1}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \le \dfrac{1}{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{8{x^2}-3x = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le \dfrac{1}{3},}\\{\left[ {\begin{array}{*{20}{c}}{x = 0,}\\{x = \dfrac{3}{8}\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 0.\) Ответ: 0.