1/
\(x+y=z+t\Rightarrow t=x+y-z\)
\(\Rightarrow t^2=\left(x+y-z\right)^2=x^2+y^2+z^2+2xy-2xz-2yz\)
Thay vào
\(B=x^2+y^2+z^2+x^2+y^2+z^2+2xy-2xz-2yz\)
\(B=x^2+2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2\)
\(B=\left(x+y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\) (đpcm)
2/
\(A=x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}+\dfrac{3y^2}{4}-\dfrac{3y}{2}-\dfrac{9}{4}\)
\(\Leftrightarrow A=\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)-3\)
\(\Leftrightarrow A=\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(\left\{{}\begin{matrix}y-1=0\\x+\dfrac{y}{2}-\dfrac{3}{2}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
b/ Nhận thấy \(x=1\) không phải là nghiệm
\(y\left(x-1\right)=x^3-x^2+2\)
\(\Leftrightarrow y=\dfrac{x^3-x^2+2}{x-1}=x^2+\dfrac{2}{x-1}\)
Do \(x;y\) nguyên \(\Rightarrow\dfrac{2}{x-1}\) nguyên
\(\Rightarrow x-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(x-1=-2\Rightarrow x=-1\Rightarrow y=0\)
\(x-1=-1\Rightarrow x=0\Rightarrow y=-2\)
\(x-1=1\Rightarrow x=2\Rightarrow y=6\)
\(x-1=2\Rightarrow x=3\Rightarrow y=10\)
Vậy pt đã cho có 4 cặp nghiệm:
\(\left(x;y\right)=\left(-1;0\right);\left(0;-2\right);\left(2;6\right);\left(3;10\right)\)
