Question

tìm các số nguyên x,y,z sao cho:

x²+y²+z²+3<xy+3y+2z

Answer 1

\(4x^2+4y^2+4z^2-4xy-12y-8z+12< 0\)

\(\Leftrightarrow\left(2x-y\right)^2+3\left(y-2\right)^2< 4-4\left(z-1\right)^2\)

Do \(\left(2x-y\right)^2+3\left(y-2\right)^2\Rightarrow4-4\left(z-1\right)^2>0\)

\(\Rightarrow\left(z-1\right)^2< 1\Rightarrow z-1=0\Rightarrow z=1\)

\(\Rightarrow\left(2x-y\right)^2< 3-3\left(y-2\right)^2\)

Tương tự ta có \(3-3\left(y-2\right)^2>0\Rightarrow y-2=0\Rightarrow y=2\)

\(\left(2x-2\right)^2< 3\Rightarrow\left(x-1\right)^2< \frac{3}{4}\)

\(\Rightarrow x-1=0\Rightarrow x=1\)