Đề thi học kì 1 Toán 9 - Đề số 5

$\begin{array}{l}\left\{ {\begin{array}{llllllllllllllllllll}{ - 2x - y = - 1}\\{ - x + 2y = - 1}\end{array}} \right.\\\left\{ {\begin{array}{llllllllllllllllllll}{ - 4x - 2y = - 2}\\{ - x + 2y = - 1}\end{array}} \right.\\\left\{ {$$\begin{array}{llllllllllllllllllll}{ - 5x = - 3}\\{ - x + 2y = - 1}\end{array}$} \right.\\\left\{ {\begin{array}{llllllllllllllllllll}{x = \frac{3}{5}}\\{ - \frac{3}{5} + 2y = - 1}\end{array}} \right.\\\left\{ {$\begin{array}{llllllllllllllllllll}{x = \frac{2}{5}}\\{y = \frac{{ - 1}}{5}}\end{array}$} \right.\end{array}

Vậy hệ phương trình có nghiệm duy nhất $\left( {x;y} \right) = \left( {\frac{2}{5};\frac{{ - 1}}{5}} \right)$.

Đáp án A