Question

Cho x,y,z là 3 số thỏa mãn đk: 

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

Tính gtrị của \(S=\left(x-4\right)^{2023}+\left(y-4\right)^{2025}+\left(z-4\right)^{2027}.\)

Answer 1

\(GT\Leftrightarrow\left(4x^2+y^2+z^2-2.2x.y-2.2x.z+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

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

Xét thấy \(VT\)\(\ge\)\(0\) với mọi x,y,z

Dấu "=" xảy ra \(\Leftrightarrow\)\(2x-y-z=0;y-3=0;z-5=0\)

\(\Leftrightarrow\)\(x=4;y=3;z=5 \)

Thay x,y,z vào S ta được: \(S=(-1)^{2025}+1^{2027}=0\)

Vậy S=0

Answer 2

\(Pt\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y-z=0\\y-3=0\\z=5=0\end{matrix}\right.\)

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

\(S=\left(x-4\right)^{2023}+\left(y-4\right)^{2025}+\left(z-4\right)^{2027}\)

\(\Rightarrow S=\left(4-4\right)^{2023}+\left(3-4\right)^{2025}+\left(5-4\right)^{2027}\)

\(\Rightarrow S=2\)

Answer 3

Sửa lại \(S=0\)