Question

Cho \(\hept{\begin{cases}x+y=a+b\\x^2+y^2=a^2+b^2\end{cases}}\)CMR \(\forall n\inℤ\)thì \(x^n+y^n=a^n+b^n\)

Answer 1

\(\hept{\begin{cases}x+y=a+b\\x^2+y^2=a^2+b^2\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=a+b\\x^2-a^2=b^2-y^2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x-a=b-y\\\left(x-a\right)\left(x+a\right)=\left(y-b\right)\left(y+b\right)\end{cases}}\) (1)

Nếu \(x=a;y=b\Rightarrow x^n+y^n=a^n+b^n\)

Nếu \(x\ne a;x\ne b\) Từ \(\left(1\right)\Rightarrow x+a=-y-b\Rightarrow x+y=-a-b\)

Mà \(x+y=a+b\Rightarrow-a-b=a+b\Leftrightarrow2\left(a+b\right)=0\Rightarrow\hept{\begin{cases}a=-b\\x=-y\end{cases}}\)

\(\Rightarrow x^n+y^n=a^n+b^n=0\)