tìm x,y Ix+3y-1I+(2y-1/2)^2000=0

Đáp án:

\(\,\left( {x;\,\,y} \right) = \left( {\frac{1}{4};\,\,\frac{1}{4}} \right)\)

Giải thích các bước giải: \[\begin{array}{l} \left| {x + 3y - 1} \right| + {\left( {2y - \frac{1}{2}} \right)^{2000}} = 0\,\,\,\left( * \right)\\ Vi\,\,\left\{ \begin{array}{l} \left| {x + 3y - 1} \right| \ge 0\,\,\,\forall x,\,\,y\\ {\left( {2y - \frac{1}{2}} \right)^{2000}} \ge 0\,\,\,\forall y \end{array} \right.\\ \Leftrightarrow \left( * \right) \Leftrightarrow \left\{ \begin{array}{l} \left| {x + 3y - 1} \right| = 0\\ {\left( {2y - \frac{1}{2}} \right)^{2000}} = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + 3y - 1 = 0\\ 2y - \frac{1}{2} = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = - 3y + 1\\ y = \frac{1}{4} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = - \frac{3}{4} + 1 = \frac{1}{4}\\ y = \frac{1}{4} \end{array} \right..\\ Vay\,\,\,\left( {x;\,\,y} \right) = \left( {\frac{1}{4};\,\,\frac{1}{4}} \right). \end{array}\]