Question

Chứng minh rằng:

Nếu \(\left(x+y+z\right)^2=x^2+y^2+z^2\) thì \(xy+yz+zx=0\)

Answer 1

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

<=>        \(x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2=0\)

<=>         \(2xy+2xz+2yz=0\)

<=>          \(2.\left(xy+xz+yz\right)=0\)

<=>           \(xy+xz+yz=0\)

Vậy_

Answer 2

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

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

\(\Leftrightarrow2\left(xy+xz+yz\right)=0\)

\(xy+xz+yz=0\left(đpcm\right)\)