\(y=\frac{x^2+1}{x^2-x+1}=\frac{\left(2x^2-2x+2\right)-x^2+2x-1}{x^2-x+1}=2-\frac{\left(x-1\right)^2}{x^2-x+1}\Rightarrow GTLNcuayla:2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Nếu bạn học \(\Delta\)r thì
\(y=\frac{x^2+1}{x^2-x+1}\Leftrightarrow x^2+1=x^2y-xy+y\)
\(\Leftrightarrow x^2\left(y-1\right)-xy+y-1=0\)
Coi pt trên là pt bậc hai ẩn x tham số y ta có:
\(\Delta=y^2-4\left(y-1\right)^2=y^2-4y^2+8y-4=-3y^2+8y-4\)
Cần \(-3y^2+8y-4\ge0\)
\(\Leftrightarrow3y^2-8y+4\le0\)
\(\Leftrightarrow3y^2-6y-2y+4\le0\)
\(\Leftrightarrow\left(3y-2\right)\left(y-2\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\le2\\y\ge\frac{3}{2}\end{matrix}\right.\)
Vậy MAX = 2<=>x=
MIN=3/2<=>x=