\(y=\dfrac{2x+2}{x^2+2x+2}\)
\(\Leftrightarrow yx^2+2xy+2y=2x+2\)
\(\Leftrightarrow yx^2+2xy+2y-2x-2=0\)
\(\Leftrightarrow yx^2+2x\left(y-1\right)+2\left(y-1\right)=0\)
\(\Delta_x=\left[2\left(y-1\right)\right]^2-8\left(y-1\right)\)
\(=4y^2-8y+4-8y+8\)
\(=4y^2-16y+12\)
\(\Leftrightarrow1\le y\le3\)
`@`\(Min_y=1\) khi \(\dfrac{2x+2}{x^2+2x+2}=1\)
\(\Leftrightarrow2x+2=x^2+2x+2\)
\(\Leftrightarrow x=0\)
`@`\(Max_y=3\) khi \(\dfrac{2x+2}{x^2+2x+2}=3\)
\(\Leftrightarrow2x+2=3x^2+6x+6\)
\(\Leftrightarrow3x^2+4x+4=0\) ( vô lý )
Vậy \(Min_y=1\) khi \(x=0\)
( Cho tớ tìm Min thôi nha chứ tớ tìm max hongg ra:"(( )
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