1) \(A=-x^2-y^2+4x-4y+2\)
\(\Leftrightarrow A=-x^2+4x-4-y^2-4y-4+4+4+2\)
\(\Leftrightarrow A=-\left(x^2-4x+4\right)-\left(y^2+4y+4\right)+\left(4+4+2\right)\)
\(\Leftrightarrow A=-\left(x-2\right)^2-\left(y+2\right)^2+10\)
Vậy GTLN của \(A=10\) khi \(\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
2) \(A=4x^2+4x+2017\)
\(\Leftrightarrow A=4x^2+4x+1-1+2017\)
\(\Leftrightarrow A=\left(4x^2+4x+1\right)+2016\)
\(\Leftrightarrow A=\left(2x+1\right)^2+2016\)
Vậy GTNN của \(A=2016\) khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)
3) Ta có:
\(3n^3+10n^2-8⋮3n+1\)
\(\Rightarrow\left(3n^3+n^2\right)+9n^2-8⋮3n+1\)
\(\Rightarrow n^2\left(3n+1\right)+9n^2-8⋮3n+1\)
\(\Rightarrow9n^2-8⋮3n+1\)
\(\Rightarrow\left(9n^2-1\right)-7⋮3n+1\)
\(\Rightarrow\left(3n-1\right)\left(3n+1\right)-7⋮3n+1\)
\(\Rightarrow-7⋮3n+1\)
\(\Rightarrow3n+1\in U\left(7\right)=\left\{-1;1;-7;7\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}3n+1=-1\Rightarrow n=\dfrac{-2}{3}\\3n+1=1\Rightarrow n=0\\3n+1=-7\Rightarrow n=\dfrac{-8}{3}\\3n+1=7\Rightarrow n=2\end{matrix}\right.\)
Vì \(n\in Z\) \(\Rightarrow n\in\left\{0;2\right\}\)
Vậy \(n=0\) hoặc \(n=2\) thì \(3n^3+10n^2-8⋮3n+1\)
4)
a) ĐKXĐ: \(\left(x+1\right)\left(2x-6\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\2x\ne6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
Vậy \(x\ne-1\) và \(x\ne3\) thì \(A\) được xác định
b) \(A=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Leftrightarrow3x^2+3x=0\)
\(\Leftrightarrow3x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=-1\left(KTMĐK\right)\end{matrix}\right.\)
Vậy để \(A=0\) thì \(x=0\)
