Question

Cho x,y,z đôi một khác nhau thoả mãn: x³+y³+z³= 3xyz (xyz \(\ne0\))

\(T\text{ính}B=\frac{16\left(x+y\right)}{z}+\frac{3\left(y+z\right)}{x}-\frac{2038\left(x+z\right)}{y}\)

Answer 1

(x+y)^3 - 3xy(x+y) + z^3 - 3xyz = 0

(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0

(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0

(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0

Suy ra x+y+z =0 

x+y = -z

y+z = -x

x+z = -y

B = -16 + (-3) +2038 = 2019

Answer 2

Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)

+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)

\(=-16-3+2038=2019\)

+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)

\(=32+6-4076=-4038\)

Answer 3

Cho em hỏi chút ạ, trường hợp  x=y=z suy ra ntn ạ?

Answer 4

\(x^2+y^2+z^2=zy+yz+zx\)(chắc em hiểu chỗ này)

\(2x^2+2y^2+2z^2=2zy+2yz+2zx\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

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

Answer 5

Cảm ơn mn :DD