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

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1. Đáp án:

    

   Giải thích các bước giải:

       x³+y³+z³=3xyz

   ⇔(x+y)³+z³-3xy(x+y)-3xyz=0

   ⇔(x+y+z)[(x+y)²-(x+y)z+z²)-3xy(x+y+z)=0

   ⇔(x+y+z)(x²+y²+z²-xy-xz-yz)=0

   ⇔(x+y+z)[(x-y)²+(x-z)²+(y-z)²]=0

   Nếu x+y+z=0

   P=$\frac{xyz}{-xyz}$ 

     =-1

   Nếu (x-y)²+(x-z)²+(y-z)²=0

   ⇒x=y=z

   P= $\frac{x^{3}}{8x^{3}}$ 

     =$\frac{1}{8}$ 

    

   Trả lời

2. `x^3+y^3+z^3=3xyz`

   `<=>x^3+y^3+z^3-3xyz=0`

   `<=>(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0`

   TH1:`x^2+y^2+z^2-xy-yz-xy=0`

   `<=>1/2 (2x^2+2y^2+2z^2-2xy-2yz-2xz)=0`

   `<=>1/2[(x-y)^2 +(y-z)^2+(x-z)^2]=0`

   `<=>(x-y)^2 +(y-2)^2 +(x-z)^2=0`

   `<=>`$\begin{cases}(x-y)^2=0\\(y-z)^2=0\\(x-z)^2=0\end{cases}$

   `<=>x=y=z`

   `=>P=(xyz)/[(x+y)(y+z)(x+z)] =1/8`

   TH2:`x+y+z=0`

   `=>P=(xyz)/[(x+y)(y+z)(x+z)]=(xyz)/[(-z).(-x).(-y)] =-1`

    

   Trả lời