a, $\frac{(x³+y³+z³-3yz)}{(x²+y²+z²-xy-zx)}$
Xét tử số:
x³+y³+z³−3xyz=(x+y)³−3xy(x+y)+z³−3xyz
=[(x+y)³+z³]−3xy(x+y+z)
=(x+y+z)[(x+y)²−(x+y)z+z²]−3xy(x+y+z)
=(x+y+z)[(x+y)²−(x+y)z+z²−3xy]
=(x+y+z)(x²+y²+z²−xy−yz−xz)
Do đó:
$\frac{x³+y³+z³−3xyzx}{x²+y²+z²−xy−yz−xz}$ =$\frac{(x+y+z)(x²+y²+z²−xy−yz−xz)}{x²+y²+z²−xy−yz−xz}$ =x+y+z
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