a, Ta có: \(\left\{{}\begin{matrix}-3x^2\le0\\-5\left|y+1\right|\le0\end{matrix}\right.\Rightarrow-3x^2-5\left|y+1\right|\le0\)
\(\Rightarrow A=-3x^2-5\left|y+1\right|+3\le3\)
Dấu " = " khi \(\left\{{}\begin{matrix}3x^2=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\)
Vậy \(MAX_A=3\) khi x = 0; y = -1
b) Ta có: \(B=-x^2-2x+7=-\left(x^2+2x-7\right)\)
\(=-\left(x^2+2x+1-8\right)\)
\(=-\left[\left(x+1\right)^2-8\right]\)
\(=-\left(x+1\right)^2+8\)
Mà \(-\left(x+1\right)^2\le0\Rightarrow B=-\left(x+1\right)^2+8\le8\)
Dấu " = " khi \(x+1=0\Rightarrow x=-1\)
Vậy \(MAX_B=8\) khi x = -1
c) \(C=\dfrac{x^2-2x+2005}{x^2-2x+2000}=\dfrac{x^2-2x+1+2004}{x^2-2x+1+1999}\)
\(=\dfrac{\left(x-1\right)^2+2004}{\left(x-1\right)^2+1999}=1+\dfrac{5}{\left(x-1\right)^2+1999}\)
Vì \(\left(x-1\right)^2+1999\ge0\) nên để \(\dfrac{5}{\left(x-1\right)^2+1999}\) lớn nhất thì \(\left(x-1\right)^2+1999\) nhỏ nhất
Ta có: \(\left(x-1\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+1999\ge1999\)
\(\Rightarrow\dfrac{5}{\left(x-1\right)^2+1999}\le\dfrac{5}{1999}\)
\(\Rightarrow C=1+\dfrac{5}{\left(x-1\right)^2+1999}\le1+\dfrac{5}{1999}=\dfrac{2004}{1999}\)
Dấu " = " khi \(x-1=0\Rightarrow x=1\)
Vậy \(MAX_C=\dfrac{2004}{1999}\) khi x = 1
Xin lỗi nhưng vì o biết nên mình dùng [ ] thay cho GTTĐ nhé
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