\(yx^2+y=x+m\Leftrightarrow yx^2-x+y-m=0\)
\(\Delta=1-4y\left(y-m\right)=-4y^2+4my+1\ge0\)
\(\Leftrightarrow-4y^2+4ym-m^2+m^2+1\ge0\)
\(\Leftrightarrow m^2+1\ge\left(2y-m\right)^2\)
\(\Rightarrow-\sqrt{m^2+1}\le2y-m\le\sqrt{m^2+1}\)
\(\Rightarrow\frac{m-\sqrt{m^2+1}}{2}\le y\le\frac{m+\sqrt{m^2+1}}{2}\)
\(y_{max}=3\Leftrightarrow\frac{m+\sqrt{m^2+1}}{2}=3\)
\(\Leftrightarrow\sqrt{m^2+1}=6-m\) (\(m\le6\))
\(\Leftrightarrow m^2+1=m^2-12m+36\Rightarrow m=\frac{35}{12}\)
Ví dụ: Cho \(\overrightarrow{AB}=\left(1;-2\right)\) ; \(\overrightarrow{CD}=\left(2x;1-x\right)\)
Tìm x để \(\left|\overrightarrow{AB}-3\overrightarrow{CD}\right|\) đạt min
\(\overrightarrow{AB}-3\overrightarrow{CD}=\left(1;-2\right)-\left(6x;3-3x\right)=\left(1-6x;3x-5\right)\)
\(\Rightarrow T=\left|\overrightarrow{AB}-3\overrightarrow{CD}\right|=\sqrt{\left(1-6x\right)^2+\left(3x-5\right)^2}\)
\(=\sqrt{45x^2-42x+26}=\sqrt{45\left(x-\frac{7}{15}\right)^2+\frac{81}{5}}\ge\sqrt{\frac{81}{5}}\)
Dấu "=" xảy ra khi \(x=\frac{7}{15}\)
Đó, cách làm kiểu thế này
