toan so 8 kho day

C

cry_with_me

oài oài :(

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

$\rightarrow \dfrac{xy + yz + xz}{xyz} = 0$

$\rightarrow xy + yz + xz =0$

$\rightarrow yz=-xy -xz$

$x^2 + 2yz = x^2 + yz - xy - xz = (x-y)(x-z)$

Tương tự : $y^2 + 2xz = (y-x)(y-z)$

$z^2 + 2xy = (z-x)(z-y)$

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

$\fbox{KQ:A=1}$
 
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