Cho $x^{3}$ +$y^{3}$+ $z^{3}$ = 3xyz. Rút gọn P= $\frac{xyz}{(x+y)(y+z)(z+x)}$

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1. Đáp án:

   \(\left[\begin{array}{l}P=-1\\P=\dfrac{1}{8}\end{array}\right.\)

   Giải thích các bước giải:

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

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

   $\leftrightarrow (x+y)^3-3x^2y-3xy^2+z^3-3xyz=0$

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

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

   $\leftrightarrow (x+y+z)(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)$

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

   $\leftrightarrow 2(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$

   $\leftrightarrow (x+y+z)[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)]=0$

   $\leftrightarrow (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]=0$

   $\leftrightarrow $\(\left[\begin{array}{l}x+y+z=0\\(x-y)^2+(y-z)^2+(x-z)^2=0\end{array}\right.\) $\leftrightarrow $\(\left[\begin{array}{l}\begin{cases}x+y=-z\\y+z=-x(1)\\z+x=-y\\\end{cases}\\x=y=z(2)\end{array}\right.\)

   Xét $(1):$

   $P=\dfrac{xyz}{(x+y)(y+z)(z+x)}=\dfrac{xyz}{(-z)(-x)(-y)}=-1$

   Xét $(2):$

   $P=\dfrac{xyz}{(x+y)(y+z)(x+z)}=\dfrac{x^3}{8x^3}=\dfrac{1}{8}$

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2. Đáp án: $P\in\{-1,\dfrac18\}$

   Giải thích các bước giải:

   Ta có:

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

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

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

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

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

   $\to (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$

   $\to (x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)=0$

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

   Nếu $x+y+z=0$

   $\to x+y=-z, y+z=-x, z+x=-y$

   $\to P=\dfrac{xyz}{(-z)\cdot (-x)\cdot (-y)}=-1$

   Nếu $(x-y)^2+(y-z)^2+(z-x)^2=0$

   $\to x=y=z$

   $\to P=\dfrac{x^3}{8x^3}=\dfrac18$

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