Lời giải:
\(\frac{x^2+y^2-z^2}{2xy}+\frac{y^2+z^2-x^2}{2yz}+\frac{x^2+z^2-y^2}{2xz}=1\)
\(\Leftrightarrow \frac{x^2+y^2-z^2}{2xy}+1+\frac{y^2+z^2-x^2}{2yz}-1+\frac{x^2+z^2-y^2}{2xz}-1=0\)
\(\Leftrightarrow \frac{(x+y-z)(x+y+z)}{2xy}+\frac{(y-z-x)(y-z+x)}{2yz}+\frac{(x-z-y)(x-z+y)}{2xz}=0\)
\(\Leftrightarrow (x+y-z)\left[\frac{x+y+z}{2xy}+\frac{y-z-x}{2yz}+\frac{x-z-y}{2xz}\right]=0\)
\(\Leftrightarrow (x+y-z)(xz+yz+z^2+xy-zx-x^2+xy-zy-y^2)=0\)
\(\Leftrightarrow (x+y-z)[z^2-(x-y)^2]=0\Leftrightarrow (x+y-z)(z-x+y)(x+z-y)=0\)
Nếu $x+y-z=0$ thì:
\(\frac{x^2+y^2-z^2}{2xy}=\frac{(x+y)^2-z^2-2xy}{2xy}=-1\); \(\frac{y^2+z^2-x^2}{2yz}=\frac{z(y-x)+z^2}{2yz}=\frac{y-x+z}{2y}=\frac{y-x+y+x}{2y}=1\)
\(\frac{x^2+z^2-y^2}{2xz}=1-(-1)-1=1\)
Ta có đpcm.
Các TH còn lại tương tự.
Vậy........
