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\(x^6+\left(y^6+15y^4+75y^2+125\right)+z^3-3x^2y^2z-15x^2z=0\)
\(\Leftrightarrow x^6+\left(y^2+5\right)^3+z^3=3x^2\left(y^2+5\right)z\)
Ta có:
\(x^6+\left(y^2+5\right)^3+z^3\ge3\sqrt[3]{x^6\left(y^2+5\right)^3z^3}=3x^2\left(y^2+5\right)z\)
Đẳng thức xảy ra khi và chỉ khi:
\(x^2=y^2+5=z\)
Từ \(x^2=y^2+5\Rightarrow\left(x-y\right)\left(x+y\right)=5\)
\(\Rightarrow\left(x;y\right)=\left(3;2\right)\Rightarrow z=9\)
Vậy có đúng 1 bộ số nguyên dương thỏa mãn pt:
\(\left(x;y;z\right)=\left(3;2;9\right)\)
\(x^2+x+13=y^2\\ \Leftrightarrow x^2-y^2+x+13=0\\ \Leftrightarrow4x^2-4y^2+4x+52=0\\ \Leftrightarrow\left(2x+1\right)^2-4y^2=51\\ \Leftrightarrow\left(2x+1-2y\right)\left(2x+1+2y\right)=51=51\cdot1=17\cdot3\left(x,y>0\right)\)
Tới đây giải ra các trường hợp thui
\(x^2-25=y\left(y+6\right)\)
\(\Leftrightarrow x^2-25=y^2+6y\)
\(\Leftrightarrow x^2-25-y^2-6y=0\)
\(\Leftrightarrow x^2-\left(y^2+6y+9\right)-16=0\)
\(\Leftrightarrow x^2-\left(y+3\right)^2=16\)
\(\Leftrightarrow\left(x+y+3\right)\left(x-y-3\right)=16\)
\(\Leftrightarrow\left(x+y+3\right);\left(x-y-3\right)\in\left\{-1;1;-2;2;-4;4;-8;8;-16;16\right\}\)
Ta giải các hệ phương trình sau :
1) \(\left\{{}\begin{matrix}x+y+3=-1\\x-y-3=-16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-4\\x-y=-15\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-11\left(loại\right)\\x-y=-15\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x+y+3=1\\x-y-3=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-2\\x-y=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=17\left(loại\right)\\x-y=19\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}x+y+3=2\\x-y-3=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x-y=11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x-y=11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=-6\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}x+y+3=-2\\x-y-3=-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-5\\x-y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-10\\x-y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=0\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}x+y+3=-4\\x-y-3=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-7\\x-y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-6\\x-y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}x+y+3=4\\x-y-3=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=8\\x-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}x+y+3=-8\\x-y-3=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-11\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-10\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-6\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}x+y+3=8\\x-y-3=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\x-y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=10\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=0\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}x+y+3=-16\\x-y-3=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-19\\x-y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-17\left(loại\right)\\x-y=2\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x+y+3=16\\x-y-3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=15\\x-y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=19\left(loại\right)\\x-y=4\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(5;-6\right);\left(-5;0\right);\left(-3;-2\right);\left(4;-3\right);\left(-5;-6\right);\left(5;0\right)\right\}\)