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.

Tên của mày là Tôm

bài này cũng khó đấy!

Câu c,

Đặt a/2009=b/2010=c/2011=k

=>a=2009.k (1)

   b=2010.k (2) 

   c=2011.k (3)

Thay (1),(2),(3) vào biểu thức:

M=4.(a-b)(b-c)-(c-a)^2 ,ta được:

M=4.(2009.k-2010.k)(2010.k-2011.k)-(2011.k-2009.k)^2

M=4.(-k).(-k)-(2k)^2

M=4.k^2-4.k^2

M=0.

PT đã cho suy ra thành

\(\left(\frac{x^{2010}}{a^2+b^2+c^2+d^2}-\frac{x^{2010}}{a^2}\right)+\left(\frac{y^{2010}}{a^2+b^2+c^2+d^2}-\frac{y^{2010}}{b^2}\right)+\left(\frac{z^{2010}}{a^2+b^2+c^2+d^2}-\frac{z^{2010}}{c^2}\right)\)

\(+\left(\frac{t^{2010}}{a^2+b^2+c^2+d^2}-\frac{t^{2010}}{d^2}\right)=0\)

\(=>x^{2010}\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+\left(tương\right)Tựnha=0\)

Do

\(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\)

máy cái bạn tự suy ra cx thế

\(=>x^{2010}=y^{2010}=z^{2010}=t^{2010}=0=>x=y=z=t=0\)

ta có 

\(T=x^{2011}+y^{2011}+z^{2011}+t^{2011}=0+0+0+0=0\)

Ta có:

\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)

<=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)

\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)=0\)(1)

Lại có: \(x^{2010};y^{2010};z^{2010};t^{2010}\ge0;\forall x,y,z,t\)

và với mọi a; b ; c ; d khác 0 có:

\(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)

\(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);

\(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\);

\(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}>0\)

=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)

\(y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)

\(z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)

\(t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)

=> \(x^{2010}\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+y^{2010}\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\)

\(+z^{2010}\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)+t^{2010}\left(\frac{1}{d^2}-\frac{1}{a^2+b^2+c^2+d^2}\right)\ge0\)

Như vậy (1) xảy ra<=> \(x^{2010}=y^{2010}=z^{2010}=t^{2010}=0\)

<=> x = y = z = t = 0

Thay vào T ta có : T = 0

\(\frac{x^{2010}+y^{2010}+z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)

=> \(\frac{x^{2010}}{a^2+b^2+c^2+d^2}+\frac{y^{2010}}{a^2+b^2+c^2+d^2}+\frac{z^{2010}}{a^2+b^2+c^2+d^2}+\frac{t^{2010}}{a^2+b^2+c^2+d^2}\)\(=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)

=> \(\left(\frac{x^{2010}}{a^2+b^2+c^2 +d^2}-\frac{x^{2010}}{a^2}\right)+\left(\frac{y^{2010}}{a^2+b^2+c^2+d^2}-\frac{y^{2010}}{b^2}\right)+\left(\frac{z^{2010}}{a^2+b^2+c^2+d^2}-\frac{z^{2010}}{c^2}\right)\)\(+\left(\frac{t^{2010}}{a^2+b^2+c^2+d^2}-\frac{t^{2010}}{d^2}\right)=0\)

=> \(x^{2010}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+y^{2010}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\right)\)\(+z^{2010}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\right)+t^{2010}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\right)=0\)

Do \(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\) ;  \(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\ne0\); \(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\ne0\)và \(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\ne0\)

\(\Rightarrow\hept{\begin{cases}x^{2010}=0\\y^{2010}=0\\z^{2010}=0;t^{2010}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\y=0\\z=0;t=0\end{cases}}\) 

Ta có : 

T = x²⁰¹¹+y²⁰¹¹ + z²⁰¹¹ + t²⁰¹¹ 

=> T = 0²⁰¹¹ + 0²⁰¹¹ + 0²⁰¹¹+ 0²⁰¹¹ 

          = 0 + 0 + 0+ 0

           = 0