Question

cho \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) tínhA=\(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)

Answer 1

Áp dụng công thức a³+b³+c³=3abc

Bài làm

Đặt \(\dfrac{1}{x}\)= a, \(\dfrac{1}{y}\)= b, \(\dfrac{1}{z}\)= c

vì a+b+c = 0 nên a³+b³+c³=3abc

S= \(\dfrac{yz}{x^2}\)+ \(\dfrac{xz}{y^2}\)+ \(\dfrac{xy}{z^{ }2}\)

=\(\dfrac{xyz}{x^{ }3}\)+\(\dfrac{xyz}{y^{ }3}\)+\(\dfrac{xyz}{z^{ }3}\) = xyz(\(\dfrac{1}{x^3}\)+\(\dfrac{1}{y^{ }3}\)+\(\dfrac{1}{z^{ }3}\))

= xyz ( a³+b³+c^{3 )}

^{= xyz \(\times\)}3abc = xyz \(\times\) \(\dfrac{3}{xyz}\) = 3