Question

Cho 4x² + 2y² + 2z² - 4xy - 4xz + 2yz - 6y - 10z + 34 = 0 .

Tính S = ( x - 4 )²⁰¹⁸ + ( y - 4 )²⁰¹⁹ + ( z - 4 )²⁰²⁰

Answer 1

Lời giải:
Ta có:

\(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)

\(\Leftrightarrow (4x^2-4xy+y^2)+2z^2+y^2-2z(2x-y)-6y-10z+34=0\)

\(\Leftrightarrow (2x-y)^2-2z(2x-y)+z^2+(y^2-6y+9)+(z^2-10z+25)=0\)

\(\Leftrightarrow (2x-y-z)^2+(y-3)^2+(z-5)^2=0\)

Vì \((2x-y-z)^2; (y-3)^2; (z-5)^2\geq 0, \forall x,y,z\). Do đó để \((2x-y-z)^2+(y-3)^2+(z-5)^2=0\) thì:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\)

\(\Rightarrow \left\{\begin{matrix} x=4\\ y=3\\ z=5\end{matrix}\right.\)

Khi đó:

\(S=(4-4)^{2018}+(3-4)^{2019}+(5-4)^{2020}=0+(-1)+1=0\)