Question

Cho x,y,z thỏa mãn \(x^2-2y=-1\); \(y^2+1=2z\) ; \(2z^2=4x-2\)

Tính \(x^{2015}+y^{2015}+z^{2015}\)

Answer 1

Ta có: \(x^2-2y=-1\) \(\Leftrightarrow\) \(x^2-2y+1=0\) (1)

\(y^2+1=2z\) \(\Leftrightarrow y^2-2z+1=0\) (2)

\(2z^2=4x-2\) \(\Leftrightarrow2z^2-4x+2=0\)(3)

Cộng (1)(2)(3) theo vế:

\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)

=> x-1=0; y-1=0; z-1=0

=>x=y=z

=>\(x^{2015}+y^{2015}+z^{2015}=1+1+1=3\)(đpcm)

Answer 2

x^2-2y=-1=>x^2-2y+1=0

y^2+1=2z=>y^2-2z+1=0

2z^2=4x-2=>z^2-2x+1=0

cộng vế với vế của 3 pt

ta có x^2-2y+1+y^2-2z+1+z^2-2x+1=0

=>(x-1)^2+(y-1)^2+(z-1)^2=0

=>x-1=0; y-1=0; z-1=0;

=>x=y=z=1

=>x^2015+y^2015+z^2015=3