Question

Answer 1

+) Thay \(A\left(-1;\dfrac{1}{3}\right)\) vào \(\left(P\right)y=\dfrac{1}{3}x^2:\)

\(\dfrac{1}{3}.\left(-1\right)^2=\dfrac{1}{3}.\)

\(\Leftrightarrow\dfrac{1}{3}=\dfrac{1}{3}\) (luôn đúng).

\(\Rightarrow\) \(A\left(-1;\dfrac{1}{3}\right)\) thuộc \(\left(P\right)y=\dfrac{1}{3}x^2.\)

+) Thay \(\text{​​}\text{​​}B\left(-\dfrac{1}{2};\dfrac{1}{6}\right)\) vào \(\left(P\right)y=\dfrac{1}{3}x^2:\)

\(\dfrac{1}{3}.\left(\dfrac{-1}{2}\right)^2=\dfrac{1}{6}.\)\(\Leftrightarrow\dfrac{1}{3}.\dfrac{1}{4}=\dfrac{1}{6}.\)\(\Leftrightarrow\dfrac{1}{12}=\dfrac{1}{6}\) (vô lý).\(\Rightarrow\) \(\text{​​}\text{​​}B\left(-\dfrac{1}{2};\dfrac{1}{6}\right)\) không thuộc \(\left(P\right)y=\dfrac{1}{3}x^2.\)