Đáp án:
a. \(\left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right. \to \left[ \begin{array}{l}
y = - 1\\
y = - 4
\end{array} \right.\)
Giải thích các bước giải:
Phương trình tọa độ giao điểm
\(\begin{array}{l}
a. - {x^2} = x - 2\\
\to {x^2} + x - 2 = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right. \to \left[ \begin{array}{l}
y = - 1\\
y = - 4
\end{array} \right.\\
b.\frac{{{x^2}}}{4} = - x - 1\\
\to \frac{{{x^2}}}{4} + x + 1 = 0\\
\to x = - 2 \to y = 1\\
c. - \frac{{{x^2}}}{2} = 3x + 4\\
\to - \frac{{{x^2}}}{2} - 3x - 4 = 0\\
\to \left[ \begin{array}{l}
x = - 2\\
x = - 4
\end{array} \right. \to \left[ \begin{array}{l}
y = - 2\\
y = - 8
\end{array} \right.\\
d.\frac{{{x^2}}}{4} = - \frac{1}{2}x\\
\to \frac{{{x^2}}}{4} + \frac{1}{2}x = 0\\
\to \frac{1}{2}x\left( {\frac{1}{2}x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right. \to \left[ \begin{array}{l}
y = 0\\
y = 1
\end{array} \right.
\end{array}\)
