Question

Answer 1

cái này khó hơn bình thg

Answer 2

\(\Leftrightarrow x^2+2y^2+5z^2+1=2xy+4yz+2z\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-4yz+4z^2+z^2-2z+1=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-2z\right)^2+\left(z-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-2z\right)^2=0\\\left(z-1\right)^2=0\end{matrix}\right.\)

Xét z-1=0 => z=1

Thế z=1 và \(\left(y-2z\right)^2=0\) => y=2

Thế y=2 vào \(\left(x-y\right)^2=0\) => x=2

Vậy \(\left\{{}\begin{matrix}x=2\\y=2\\z=1\end{matrix}\right.\)

 

 

Answer 3

đi thi HSG à:))