Question

cho x,y,z là các số hữu tỉ thoả mãn \(\left(x-y+z\right)^2=x^2-y^2+z^2\).CMR:

\(\left(x-y+z\right)^n=x^n-y^n+z^n\)

Answer 1

Ta có : \(\left(x-y+z\right)^2=x^2-y^2+z^2\)

\(\Leftrightarrow x^2+y^2+z^2-2xy+2xz-2yz=x^2-y^2+z^2\)

\(\Leftrightarrow2y^2-2xy+2xz-2yz=0\)

\(\Leftrightarrow2y\left(y-z\right)-2x\left(y-z\right)=0\)

\(\Leftrightarrow2\left(y-x\right)\left(y-z\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=z\end{matrix}\right.\)

Với x = y \(\Rightarrow\left(x-y+z\right)^n=z^n;x^n-y^n+z^n=z^n\)

\(\Rightarrow\left(x-y+z\right)^n=x^n-y^n+z^n\) ( 1 )

Với y = z \(\Rightarrow\left(x-y+z\right)^n=x^n;x^n-y^n+z^n=x^n\)

\(\Rightarrow\left(x-y+z\right)^n=x^n-y^n+z^n\) ( 2 )

Từ ( 1 ) ; ( 2 ) => ĐPCM