b: A(1;1) B(-2;4)
\(M\left(x;x^2\right)\)
Theo đề, ta có: MA=MB
\(\Leftrightarrow\sqrt{\left(x-1\right)^2+\left(x^2-1\right)^2}=\sqrt{\left(x+2\right)^2+\left(x^2-4\right)^2}\)
\(\Leftrightarrow x^2-2x+1+x^4-2x^2+1=x^2+4x+4+x^4-8x^2+16\)
\(\Leftrightarrow6x^2-6x-18=0\)
\(\Leftrightarrow x^2-x-3=0\)
\(\Delta=\left(-1\right)^2-4\cdot1\cdot\left(-3\right)=13>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{13}}{2}\\x_2=\dfrac{1+\sqrt{13}}{2}\end{matrix}\right.\)
Vậy: \(M\left(\dfrac{1-\sqrt{13}}{2};\dfrac{7-\sqrt{13}}{2}\right);M\left(\dfrac{1+\sqrt{13}}{2};\dfrac{7+\sqrt{13}}{2}\right)\)
