cho biểu thức `A=(4xy-z^2)/(xy+2z^2).(4yz-x^2)/(yz+2x^2).(4zx-y^2)/(xz+2y^2) `. cm rằng nếu x+y+z=0 thì A =1
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$x+y+z=0$
$=>x=-y-z$
$=> x^2=(y+z)^2=y^2+2yz+z^2$
Tương tự:
$y^2=x^2+2xz+z^2$
$z^2=x^2+2xy+y^2$
Có:
$4xy-z^2=4xy-(x^2+2xy+y^2)=2xy-x^2-y^2=-(x-y)^2$
Tương tự:
$4yz-x^2=-(y-z)^2$
$4zx-y^2=-(z-x)^2$
$xy+2z^2$
$=(-y-z)y+2z^2$
$= -y^2-yz+2z^2$
$= (z^2-y^2) + (z^2-yz)$
$=(z-y)(z+y) + z(z-y)$
$=(z-y)(2z+y)$
$=(z-y)(2z-x-z)$
$=(z-y)(z-x)$
Tương tự:
$yz+2x^2=(x-y)(x-z)$
$xz+2y^2=(y-x)(y-z)$
$=>A=\dfrac{-(x-y)^2}{(z-y)(z-x)}.\dfrac{-(y-z)^2}{(x-y)(x-z)}.\dfrac{-(z-x)^2}{(y-x)(y-z)}$
$=\dfrac{-(x-y)^2}{-(x-y)^2}.\dfrac{-(y-z)^2}{-(y-z)^2}.\dfrac{-(z-x)^2}{-(z-x)^2}$
$=1$
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