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I:
a: \(=x^2-2x+1+x^2-4x+4\)
\(=2x^2-6x+5\)
\(=2\left(x^2-3x+\dfrac{5}{2}\right)\)
\(=2\left(x^2-3x+\dfrac{9}{4}+\dfrac{1}{4}\right)\)
\(=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu = xảy ra khi x=3/2
b: \(=-4\left(x^2-2x+\dfrac{3}{4}\right)\)
\(=-4\left(x^2-2x+1-\dfrac{1}{4}\right)=-4\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
\(\Leftrightarrow x^2=\left(y+1\right)^2+12\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+1\right)=12\)
+ \(\left(x-y-1\right)+\left(x+y+1\right)=2x⋮2\)
=> \(x-y-1\) và \(x+y+1\) cùng tính chẵn lẻ
\(\left\{{}\begin{matrix}x-y-1< x+y+1\\x+y+1\ge3\end{matrix}\right.\) ( do x,y nguyên dương ) và
\(x-y-1\), \(x+y+1\) cùng tính chẵn lẻ nên chỉ xảy ra TH
+ \(\left\{{}\begin{matrix}x-y-1=2\\x+y+1=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=3\\x+y=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\) ( TM )
\(xz=y^2\Rightarrow2xz=2y^2\)
\(x^2+z^2+99=7y^2\)
\(\Rightarrow x^2+z^2+2xz+99=7y^2+2y^2\)
\(\Rightarrow\left(x+z\right)^2+99=9y^2=\left(3y\right)^2\)
\(\Rightarrow\left(x+z\right)^2-\left(3y\right)^2=-99\)
\(\Rightarrow\left(x+z+3y\right)\left(x+z-3y\right)=-99=-\left(9.11\right)=-\left(3.33\right)=-\left(99.1\right)\)
Gọi: \(x+z=a;3y=b\)
\(\Rightarrow\left(a+b\right)\left(a-b\right)=-\left(99.1\right)=-\left(3.33\right)=-\left(99.1\right)\)
Trường hợp 1: \(\left(a+b\right)\left(a-b\right)=-\left(9.11\right)\)
\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a+b=11\\a-b=-9\end{matrix}\right.\\\left\{{}\begin{matrix}a+b=9\\a-b=-11\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=1\\b=10\end{matrix}\right.\\\left\{{}\begin{matrix}a=-1\\b=10\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+z=1\\3y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x+z=-1\\3y=10\end{matrix}\right.\end{matrix}\right.\) \(\left(ktm\right)\)
Trường hợp 2: \(\left(a+b\right)\left(a-b\right)=-\left(9.11\right)\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+b=33\\a-b=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=15\\b=18\end{matrix}\right.\\\Rightarrow\left\{{}\begin{matrix}x+z=15\\y=6\Rightarrow xz=6^2=36\end{matrix}\right.\\\left\{{}\begin{matrix}a+b=3\\a-b=-33\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+z=15\\3y=18\end{matrix}\right.\\\left\{{}\begin{matrix}x=12\\y=6\\z=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+z=-15\\3y=18\end{matrix}\right.\end{matrix}\right.\)
Trường hợp 3: Không thỏa mãn
Vậy \(x=12;y=6;z=3\) hoặc \(x=3;y=6;z=12\)