do vai trò của x;y;z;t là như nhau
giả sử: [tex]0< x\leq y\leq z\leq t[/tex]
có: [tex]3.(x+y+z+t)+1=8xyzt\\\\ <=> \frac{3x}{xyzy}+\frac{3y}{xyzt}+\frac{3z}{xyzt}+\frac{3t}{xyzt}+\frac{1}{xyzt}=8\\\\ <=> \frac{3}{yzt}+\frac{3}{xzt}+\frac{3}{xyt}+\frac{3}{xyz}+\frac{1}{xyzt}=8\\\\ \leq \frac{3}{t^3}+\frac{3}{t^3}+\frac{3}{t^3}+\frac{3}{t^3}+\frac{1}{t^4}\\\\ =\frac{12t+1}{t^4}\\\\ => 8\leq \frac{12t+1}{t^4}\\\\ => 8t^4\leq 12t+1\\\\ <=> 8t^3\leq 12+\frac{1}{t}<18 (do t>0)\\\\ <=> t^3<2=> t=1\\\\ => 3.(x+y+z+1)+1=8xyz\\\\ <=> \frac{3x}{xyz}+\frac{3y}{xyz}+\frac{3z}{xyz}+\frac{4}{xyz}=8\\\\ <=> \frac{3}{yz}+\frac{3}{xz}+\frac{3}{xy}+\frac{4}{xyz}=8\\\\ \leq \frac{3}{z^2}+\frac{3}{z^2}+\frac{3}{z^2}+\frac{4}{z^3}\\\\ =\frac{9z+4}{z^3}\\\\ => 8\leq \frac{9z+4}{z^3}\\\\ => 8.z^3\leq 9z+4\\\\ => 8z^2\leq 9+\frac{4}{z}<9 (do z>0)\\\\ => z=1\\\\ => 3.(x+y+1+1)+1=8xy\\\\ <=> 3x+3y+6+1=8xy\\\\ <=> 3x+3y-8xy+7=0\\\\ <=> 8.(3x+3y-8xy+7)=0\\\\ <=> 24x+24y-64xy+56=0\\\\ <=> 8x.(3-8y)-3.(3-8y)+9+56=0\\\\ <=> (8x-3).(3-8y)=-65\\\\ <=> (8x-3).(8y-3)=65[/tex]
đến đây tìm x;y thỏa mãn x;y nguyên dương rồi kết luận phương trình thôi...