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Bài 2:
a: Theo đề, ta có:
\(\left\{{}\begin{matrix}a+b+c=0\\c=5\\\dfrac{-b}{2a}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=-5\\b=-6a\\c=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5a=-5\\b=-6a\\c=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-6\\c=5\end{matrix}\right.\)
b: Theo đề, ta có:
\(\left\{{}\begin{matrix}4a+2b+c=3\\\dfrac{-b}{2a}=3\\-\dfrac{b^2+4ac}{4a}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+2b+c=3\\b=-6a\\\left(-6a\right)^2+4ac=-16a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4a-12a+c=3\\b=-6a\\36a^2+16a+4ac=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=8a+3\\b=-6a\\36a^2+16a+4a\left(8a+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{7}{17}\\b=6\cdot\dfrac{7}{17}=\dfrac{42}{17}\\c=8\cdot\dfrac{-7}{17}+3=-\dfrac{5}{17}\end{matrix}\right.\)
\(M\in\left(d_1\right)\Rightarrow M\left(x;\dfrac{x+3}{2}\right)\)
\(\left|\overrightarrow{MA}+\overrightarrow{MB}\right|=\left|\overrightarrow{MI}\right|\) \(\left(\overrightarrow{IA}=\overrightarrow{BI}\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x_I=\dfrac{x_A+x_B}{2}=-\dfrac{1}{2}\\y_I=\dfrac{y_A+y_B}{2}=\dfrac{7}{2}\end{matrix}\right.\Rightarrow I\left(-\dfrac{1}{2};\dfrac{7}{2}\right)\)
\(\Rightarrow\left|\overrightarrow{MA}+\overrightarrow{MB}\right|_{min}\Leftrightarrow\left|\overrightarrow{MI}\right|_{min}\Leftrightarrow\overrightarrow{MI}\perp\overrightarrow{AB}\Leftrightarrow\overrightarrow{MI}.\overrightarrow{AB}=0\)
\(\Leftrightarrow\left(x_I-x_M;y_I-y_M\right).\left(x_B-x_A;y_B-y_A\right)=0\)
\(\Leftrightarrow\left(x_I-x_M\right)\left(x_B-x_A\right)+\left(y_I-y_M\right)\left(y_B-y_A\right)=0\)
\(\Leftrightarrow\left(-\dfrac{1}{2}-x\right).\left(-3\right)+\dfrac{7}{2}-\dfrac{x+3}{2}=0\Rightarrow M\left(...\right)\)
\(\left|\overrightarrow{MA}+\overrightarrow{MB}\right|=2\left|\overrightarrow{MI}\right|\) nhé, đánh thiếu, nhưng nó ko ảnh hưởng gì đến bài toán :v
Do M thuộc Oy nên \(x_M=0\Rightarrow M\left(0;y\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(1;-y\right)\\\overrightarrow{MB}=\left(0;5-y\right)\\\overrightarrow{MC}=\left(-3;-5-y\right)\end{matrix}\right.\)
\(\Rightarrow3\overrightarrow{MA}-2\overrightarrow{MB}+4\overrightarrow{MC}=\left(-9;-30-3y\right)\)
\(\Rightarrow\left|3\overrightarrow{MA}-2\overrightarrow{MB}+4\overrightarrow{MC}\right|=\sqrt{\left(-9\right)^2+\left(-30-3y\right)^2}\ge9\)
Dấu = xảy ra khi \(-30-3y=0\Rightarrow y=-10\)
\(\Rightarrow S=-10\)
Biểu thức trong trị tuyệt đối kia là vecto hay ko em nhỉ? Tức là \(\left|3MA-2MB+4MC\right|\) hay \(\left|3\overrightarrow{MA}-2\overrightarrow{MB}+4\overrightarrow{MC}\right|\)?