Hệ phương trình x^2} + {y^2} + 2( {x + y} \right) = 7y( {y - 2x}- 2x = 10}

Đáp án

4.

Giải thích

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{{x^2} + {y^2} + 2\left( {x + y} \right) = 7}\\{y\left( {y - 2x} \right) - 2x = 10}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{x^2} + {y^2} + 2x + 2y = 7}\\{{y^2} - 2xy - 2x = 10}\end{array}} \right.} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{x^2} + 2xy + 2x + 10 + 2x + 2y = 7}\\{{y^2} = 2xy + 2x + 10}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{x^2} + 2xy + 4x + 2y + 3 = 0}\\{{y^2} = 2xy + 2x + 10}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\left( {{x^2} + x} \right) + \left( {2xy + 2y} \right) + \left( {3x + 3} \right) = 0}\\{{y^2} = 2xy + 2x + 10}\end{array}} \right.\)

\( \Rightarrow \left( {x + 1} \right)\left( {x + 2y + 3} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = - 1}\\{x = - 2y - 3}\end{array}} \right.\)

+) Với \(x = - 1\) ta có: \({y^2} = - 2y + 8 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{y = 2}\\{y = - 4}\end{array}} \right.\)

+) Với \(x = - 2y - 3\) ta có:

\({y^2} = 2y\left( { - 2y - 3} \right) + 2\left( { - 2y - 3} \right) + 10 \Leftrightarrow 5{y^2} + 10y - 4 = 0 \Leftrightarrow y = \frac{{ - 5 \pm 3\sqrt 5 }}{5}\)

\( \Rightarrow x = \frac{{ - 5 \mp 6\sqrt 5 }}{5}\).