Question

CHo x,y,z đôi một khác nhau thoả mãn:

x³+y³+z³ = 3xyz (xyz khác 0)

Tính \(B=\frac{16\left(x+y\right)}{z}+\frac{3\left(y+z\right)}{x}+\frac{2038\left(x+z\right)}{y}\)

Answer 1

Lời giải:

$x^3+y^3+z^3-3xyz=0$

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

$\Leftrightarrow (x+y)^3+z^3-3xy(x+y+z)=0$

$\Leftrightarrow (x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)=0$

$\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0$

Đến đây xét 2TH:

TH1: $x+y+z=0$

\(\Rightarrow \left\{\begin{matrix} x+y=-z\\ y+z=-x\\ x+z=-y\end{matrix}\right.\)

\(\Rightarrow B=-16+(-3)+(-2038)=-2057\)

TH2: $x^2+y^2+z^2-xy-yz-xz=0$

$\Leftrightarrow \frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}=0$

$\Rightarrow (x-y)^2=(y-z)^2=(z-x)^2=0$

$\Rightarrow x=y=z$ (vô lý vì $x,y,z$ đôi một khác nhau)

Vậy.......

Answer 2

\(x^3+y^3+z^3-3xyz=0\Leftrightarrow x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz=0\)

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)

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

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