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ta có: \(\frac{x^2-yz}{a}=\frac{y^2-xz}{b}=\frac{z^2-xy}{c}\)
\(\Rightarrow\frac{a}{x^2-yz}=\frac{b}{y^2-xz}=\frac{c}{z^2-xy}\Rightarrow\frac{a^2}{\left(x^2-yz\right)^2}=\frac{b^2}{\left(y^2-xz\right)^2}=\frac{c^2}{\left(z^2-xy\right)^2}\) (1)
=> \(\frac{a}{\left(x^2-yz\right)}.\frac{a}{\left(x^2-yz\right)}=\frac{b}{y^2-xz}.\frac{c}{z^2-xy}=\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-xz\right).\left(z^2-xy\right)}\)
a^2/(x^2-yz)^2 = (a^2-bc)/[(x^2-yz)^2 - (y^2-xz)(z^2-xy)] = (a^2-bc)/[x (x^3 + y^3 + z^3 - 3xyz)] =>
(a^2-bc)/x = [a^2/(x^2 - yz)^2] * (x^3 + y^3 + z^3 - 3xyz) (2)
Thực hiện tương tự ta cũng có
(b^2-ac)/y = [b^2/(y^2 - xz)^2] * (x^3 + y^3 + z^3 - 3xyz) (3)
(c^2-ab)/z = [c^2/(z^2 - xy)^2] * (x^3 + y^3 + z^3 - 3xyz) (4)
Từ (1),(2),(3),(4) => (a^2-bc)/x = (b^2-ac)/y = (c^2-ab)/z.
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(=\left(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}\right)+\left(\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}\right)+\left(\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}\right)=0\)
\(=x^2.\frac{b^2+c^2}{a^2+b^2+c^2}+y^2.\frac{a^2+c^2}{a^2+b^2+c^2}+z^2.\frac{a^2+b^2}{a^2+b^2+c^2}=0\)
Vì \(a,b,c\ne0\) nên dấu = xảy ra khi \(x=y=z=0\)
\(\Rightarrow A=x^{2003}+y^{2003}+z^{2003}=0+0+0=0\)
Ta có:
\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+az}\left(x;y;z\ne0\right)\)
=> \(\frac{xyz}{azy+bxz=}=\frac{xyz}{xbz+xcy}=\frac{yzx}{ycx+azy}\)
=>\(zay+bxz=xbz+xyc=ycx+azy\)
\(\Rightarrow\hept{\begin{cases}za=cx\\bx=ay\end{cases}}\)
Đặt \(\frac{x}{a}=\frac{z}{c}=\frac{y}{b}=t\left(t\ne0\right)\)
=> x = at ; z = ct ; y = bt
mà\(\frac{xy}{ay+bx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\)\(\frac{atbt}{abt+bat}=\frac{a^2t^2+b^2t^2+c^2t^2}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{t}{2}=t^2\Rightarrow t=\frac{1}{2}\)
\(\Rightarrow t=\frac{1}{2}\Rightarrow\hept{\begin{cases}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{cases};\left(a,b,c\ne0\right)}\)
Ta có
\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
\(=>\frac{x^{2010}}{a^2+b^2+c^2+d^2}+\frac{y^{2010}}{a^2+b^2+c^2+d^2}+\frac{z^{2010}}{a^2+b^2+c^2+d^2}+\frac{t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
\(=>\left(\frac{x^{2010}}{a^2+b^2+c^2+d^2}-\frac{x^{2010}}{a^2}\right)+\left(\frac{y^{2010}}{a^2+b^2+c^2+d^2}-\frac{y^{2010}}{b^2}\right)+\left(\frac{z^{2010}}{a^2+b^2+c^2+d^2}-\frac{z^{2010}}{c^2}\right)+\left(\frac{t^{2010}}{a^2+b^2+c^2+d^2}-\frac{t^{2010}}{d^2}\right)=0\)
\(=>x^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+y^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\right)+z^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\right)+t^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\right)=0\)
\(Do\left\{\begin{matrix}\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\ne0\end{matrix}\right.\)
\(=>\left\{\begin{matrix}x^{2010}=0\\y^{2010}=0\\z^{2010}=0\\t^{2010}=0\end{matrix}\right.\)
\(=>\left\{\begin{matrix}x=0\\y=0\\z=0\\t=0\end{matrix}\right.\)
Ta có
\(T=x^{2011}+y^{2011}+z^{2011}+t^{2011}\)
\(=>T=0^{2011}+0^{2011}+0^{2011}+0^{2011}\\ T=0+0+0+0\\ T=0\)
(x^2+y^2+z^2)/(a^2+b^2+c^2)=
=x^2/a^2+y^2/b^2+z^2/c^2 <=>
x^2+y^2+z^2=x^2+(a^2/b^2)y^2+
+(a^2/c^2)z^2+(b^2/a^2)x^2+y^2+
+(b^2/c^2)z^2+(c^2/a^2)x^2+
+(c^2/b^2)y^2+z^2 <=>
[(b^2+c^2)/a^2]x^2+[(a^2+c^2)/b^2]y^2+
+[(a^2+b^2)/c^2]z^2 = 0 (*)
Đặt A=[(b^2+c^2)/a^2]x^2; B=[(a^2+c^2)/b^2]y^2;
và C=[(a^2+b^2)/c^2]z^2
Vì a,b,c khác 0 nên suy ra A,B,C đều không âm
Từ (*) ta có A+B+C=0
Tổng 3 số không âm bằng 0 thì cả 3 số đều phải bằng 0,tức A=B=C=0
Vì a,b,c khác 0 nên [(b^2+c^2)/c^2]>0 =>x^2=0 =>x=0
Tương tự B=C=0 =>y^2=z^2=0 => y=z=0
Vậy x^2011+y^2011+z^2011=0
Và x^2008+y^2008+z^2008=0.
