Cho $x^{3}$ + $y^{3}$ + $z^{3}$ = 3xyz. Hãy rút gọn P = $\frac{xyz}{(x + y)(y + z)(z + x}$ 10/10/2021 Bởi Valerie Cho $x^{3}$ + $y^{3}$ + $z^{3}$ = 3xyz. Hãy rút gọn P = $\frac{xyz}{(x + y)(y + z)(z + x}$
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= $\frac{x^{3}}{8x^{3}}$ =$\frac{1}{8}$ Bình luận
`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` Bình luận
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= $\frac{x^{3}}{8x^{3}}$
=$\frac{1}{8}$
`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`