Question

cho x,y,z thoản mãn: 9x²+y²+z²-36x-16y+10z= -125
Tính xy+yz+xz=?

Answer 1

Ta có:

\(9x^2+y^2+z^2-36x-16y+10z=-125\)

\(\Leftrightarrow\)  \(9x^2+y^2+z^2-36x-16y+10z+125=0\)

\(\Leftrightarrow\)  \(9x^2-36x+36+y^2-16y+64+z^2+10z+25=0\)

\(\Leftrightarrow\)  \(9\left(x-2\right)^2+\left(y-8\right)^2+\left(z+5\right)^2=0\)

Mà   \(\left(x-2\right)^2;\left(y-8\right)^2;\left(z+5\right)^2\ge0\)  với mọi   \(x;y;z\)

nên   \(\left(x-2\right)^2=0;\left(y-8\right)^2=0;\left(z+5\right)^2=0\)

\(\Leftrightarrow\)   \(x-2=0;y-8=0;z+5=0\)

\(\Leftrightarrow\)   \(x=2;y=8;z=-5\)

Vậy,   \(xy+yz+xz=-34\)