tính giá trị biểu thức

P

phamhuy20011801

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0 \leftrightarrow xy+yz+zx=0$
$x^2+2xy=x^2+xy+(-yz-zx)=(x-y)(x-z)$
Tương tự $y^2+2zx=(y-z)(y-x)$
$z^2+2xy=(z-x)(z-y)$
$\rightarrow A=\dfrac{-zy(y-z)-yz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}=\dfrac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=1$
 
Last edited by a moderator:
P

pinkylun

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$

$=>xy+yz+xz=0$

$BT<=>\dfrac{yz}{x^2+yz-xy-xz}+\dfrac{xz}{y^2+xz-xy-yz}+\dfrac{xy}{z^3+xy-yz-xz}$

$<=>\dfrac{yz}{(x-y)(x-z)}+\dfrac{xz}{(y-x)(y-z)}+\dfrac{xy}{(z-y)(z-x)}$

$<=>\dfrac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(x-z}$

Bạn nhân ra rút gọn $=\dfrac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=1đpcm$
 
You must log in or register to reply here.
Top Bottom