Cho $x,y,z$ đôi một khác nhau thỏa mãn $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
Tính $A=\frac{yz}{x^{2}+2yz}+\frac{zx}{y^{2}+2zx}+\frac{xy}{z^{2}+2xy}$
Cho $x,y,z$ đôi một khác nhau thỏa mãn $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
Tính $A=\frac{yz}{x^{2}+2yz}+\frac{zx}{y^{2}+2zx}+\frac{xy}{z^{2}+2xy}$
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Cho $x,y,z$ đôi một khác nhau thỏa mãn $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
Tính $A=\frac{yz}{x^{2}+2yz}+\frac{zx}{y^{2}+2zx}+\frac{xy}{z^{2}+2xy}$
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tu dk suy ra xy+yz+xz=0 ta co
x2 +2yz=x2 +yz-xy-xz=(x-y)(x-z)
tuong tu y2+2xz=(z-y)(x-y); z2+2xy=(y-z)(x-z)
quy dong mau la (x-y)(x-z)(y-z)
tu la yz(y-z)-xz(x-z)+xy(x-y)=yz(y-z)-xz(x-y+y-z)+xy(x-y)=z(y-z)(y-x)+x(x-y)(y-z)=(x-y)(x-z)(y-z)
suy ra A=1
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 => xy + yz + zx = 0 => xy = -yz-xz
yz = -xy-xz
xz = -xy-yz$
Khi đó
A = $\frac{yz}{x^{2}+2yz}+ \frac{zx}{y^{2}+2zx}+\frac{xy}{z^{2}+2xy}
= \frac{yz}{x^{2}+yz+yz}+ \frac{zx}{y^{2}+zx+zx}+\frac{xy}{x^{2}+xy+xy}$
= $\frac{yz}{x^{2}+yz-xy-xz}+\frac{zx}{y^{2}+zx-xy-yz}+ \frac{xy}{z^{2}+xy-zx-yz}
= \frac{yz}{\left ( x-y \right )\left ( x-z \right )}-\frac{zx}{\left ( x-y \right )\left ( y-z \right )}+\frac{xy}{\left ( y-z \right )\left ( x-z \right )}$
= $\frac{yz(y-z)-xz(x-z)+xy(x-y)}{\left ( x-y \right )\left ( y-z \right )\left ( x-z \right )}
=\frac{yz(y-z)-x^{2}z+xz^{2}+x^{2}y-xy^{2}}{\left ( x-y \right )\left ( y-z \right )\left ( x-z \right )}$
=$\frac{xy\left ( x-y \right )+x^{2}(y-z)-x\left ( y+z \right )\left ( y-z \right )}{\left ( x-y \right )\left ( y-z \right )\left ( x-z \right )}$
= $\frac{\left ( x-y \right )\left ( y-z \right )\left ( x-z \right )}{\left ( x-y \right )\left ( y-z \right )\left ( x-z \right )}=1$
Mình làm hơi tắt 1 chút nhưng cũng không khó hiểu đâu
$xy+yz+zx=0\Rightarrow x^{2}+2yz=-xy+yz-zx+x^{2}=(x-y)(x-z)$
tương tự $y^{2}+2zx=(y-z)(y-x)$
$z^{2}+2xy=(z-y)(z-x)$
$\Rightarrow A=\frac{yz}{(x-y)(x-z)}+\frac{zx}{(y-z)(y-x)}+\frac{xy}{(z-x)(z-y)}$
quy đồng ta được $A=1$
Edited by tank06536, 06-06-2015 - 09:46.
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