a)\(I=2011.\left|2x-4\right|+2012.\left(y+1\right)^2+\left(-1\right)\\ +Có:\left|2x-4\right|\ge0với\forall x\Rightarrow2011.\left|2x-4\right|\ge0\\ \left(y+1\right)^2\ge0với\forall y\Rightarrow2012.\left(y+1\right)^2\ge0\\ \Rightarrow2011.\left|2x-4\right|+2012.\left(y+1\right)^2+\left(-1\right)\ge-1\\ \Leftrightarrow I\ge-1\\ +dấu"="xảyrakhi\\ \Rightarrow\left\{{}\begin{matrix}\left|2x-4\right|=0\\\left(y+1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
vậy Imin= -1 khi x = 2, y = -1
a) \(I=2011\cdot\left|2x-4\right|+2012\cdot\left(y+1\right)^2+\left(-1\right)\)
Có: \(\left|2x-4\right|\ge0\forall x\Rightarrow2011\cdot\left|2x-4\right|\ge0\forall x\)
\(\left(y+1\right)^2\ge0\forall y\Rightarrow2012\cdot\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2011\cdot\left|2x-4\right|+2012\cdot\left(y+1\right)^2\ge0\forall x;y\)
\(\Rightarrow2011\cdot\left|2x-4\right|+2012\cdot\left(y+1\right)^2+\left(-1\right)\ge0+\left(-1\right)\forall x;y\\ \Rightarrow I\ge-1\forall x;y\\ \Rightarrow I_{min}=-1\)
\("="\Leftrightarrow\left\{{}\begin{matrix}\left|2x-4\right|=0\\\left(y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-4=0\\y+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
b) \(J=\frac{2}{\left(x+2\right)^2+3}\)
Có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+3\ge3\forall x\)
\(\Rightarrow\frac{2}{\left(x+2\right)^2+3}\le\frac{2}{3}\forall x\\ \Rightarrow J\le\frac{2}{3}\forall x\\ \Rightarrow J_{max}=\frac{2}{3}\)
\("="\Leftrightarrow\left(x+2\right)^2=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\)
b)\(J=\frac{2}{\left(x+2\right)^2+3}\)
vì J có tử =2 không đổi
⇒ để J đạt giá trị max thì mẫu phải đạt min
\(+có:\left(x+2\right)^2\ge0với\forall x\Rightarrow\left(x+2\right)^2+3\ge3\\ +dấu"="xảyrakhi\\ \left(x+2\right)^2=0\Rightarrow x=-2\\ \)
⇒mẫu đạt min =3 khi x =-2
\(J_{max}=\frac{-2}{3}khix=-2\)
