a.
\(A=\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+x^2-4x+4-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(x-2\right)^2-3\)
\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\)
Do \(\left\{{}\begin{matrix}\left(x+y+1\right)^2\ge0\\\left(x-2\right)^2\ge0\end{matrix}\right.\); \(\forall x;y\)
\(\Rightarrow A\ge-3\)
\(A_{min}=-3\) khi \(\left\{{}\begin{matrix}x+y+1=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(2;-3\right)\)
b.
Kiểm tra lại đề câu này
c.
\(C=\left(x^2+xy+\dfrac{y^2}{4}\right)-3\left(x+\dfrac{y}{2}\right)+\dfrac{3y^2}{4}-\dfrac{3y^2}{2}\)
\(=\left(x+\dfrac{y}{2}\right)^2-3\left(x+\dfrac{y}{2}\right)+\dfrac{9}{4}+\dfrac{3}{4}\left(y^2-2y+1\right)-3\)
\(=\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2-3\ge-3\)
\(C_{min}=-3\) khi \(\left\{{}\begin{matrix}x+\dfrac{y}{2}-\dfrac{3}{2}=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
d.
\(D=x\left(x+1\right)\left(x^2+x-4\right)=\left(x^2+x\right)\left(x^2+x-4\right)\)
\(=\left(x^2+x\right)^2-4\left(x^2+x\right)+4-4\)
\(=\left(x^2+x-2\right)^2-4\ge-4\)
\(D_{min}=-4\) khi \(x^2+x-2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Câu b nếu muốn vẫn có thể làm được (vấn đề là kết quả xấu quá):
\(B=\left(x^4-2x^2y^2+y^4\right)+3x^2y^2-8xy+xy\left(x^2-y^2\right)+200\)
\(=\left(x^2-y^2\right)^2+xy\left(x^2-y^2\right)+\dfrac{x^2y^2}{4}+\dfrac{11}{4}x^2y^2-8xy+200\)
\(=\left(x^2-y^2+\dfrac{xy}{2}\right)^2+\dfrac{11}{4}\left(xy-\dfrac{16}{11}\right)^2+\dfrac{2163}{11}\)
\(\Rightarrow B\ge\dfrac{2163}{11}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2-y^2+\dfrac{xy}{2}=0\\xy=\dfrac{16}{11}\end{matrix}\right.\) giải cái này ra x;y xấu quá
