Đặt $\sqrt[]{a}$=x ;$\sqrt[]{b}$=y;$\sqrt[]{c}$;z
Đặt cho đễ viết thôi chứ cx ko cần
Ta có
H= $\frac{x-y}{1+z²}$ +$\frac{y-z}{1+x²}$ +$\frac{z-x}{1+y²}$
Vì xy+yz+zx=1 (gt)
Nên H=$\frac{x-y}{xy+yz+zx+z²}$ +$\frac{y-z}{xy+yz+zx+x²}$ +$\frac{z-x}{xy+yz+zx+y²}$
=$\frac{x-y}{y.(x+z)+z(x+z)}$ +$\frac{y-z}{y(z+x)+x(z+x)}$ +$\frac{z-x}{y(y+z)+x(y+z)}$
=$\frac{x-y}{(y+z).(z+x))}$+$\frac{y-z}{(x+y)(x+z)}$+$\frac{z-x}{(x+y).(y+z)}$
=$\frac{(x-y).(x+y)}{(x+y).(y+z).(z+x)}$ +$\frac{(y-z).(y+z)}{(x+y).(y+z).(z+x)}$ +$\frac{(z-x).(z+x)}{(x+y).(y+z).(z+x)}$
=$\frac{x²-y²}{(x+y).(y+z).(z+x)}$+$\frac{y²-z²}{(x+y).(y+z).(z+x)}$+$\frac{z²-x²}{(x+y).(y+z).(z+x)}$
=$\frac{(x²-y²)+(y²-z²)+(z²-x²)}{(x+y).(y+z).(z+x)}$
=$\frac{0}{(x+y).(y+z).(z+x)}$
=0
Vậy H=0
