Question

Biết \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\). Khi đó giá trị biểu thức \(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\) là bao nhiêu ?

Answer 1

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

<=> \(\dfrac{yz}{xyz}+\dfrac{xz}{xyz}+\dfrac{xy}{xyz}=0\)

<=> yz + xz + xy = 0

=> (yz)³ + (xz)³ + (xy)³ = 3x²y²z²

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

= \(\dfrac{y^3z^3}{x^2y^2z^2}+\dfrac{x^3z^3}{x^2y^2z^2}+\dfrac{x^3y^3}{x^2y^2z^2}\)

= \(\dfrac{3x^2y^2z^2}{x^2y^2z^2}\)

= 3

Answer 2

Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)

=> \(\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\dfrac{1}{z^3}\)

<=> \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+3.\dfrac{1}{x}.\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{-1}{z^3}\)

<=> \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

<=>\(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}\)(do \(\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\))

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

Do đó ta có:

A=\(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)=\(\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz.3.\dfrac{1}{xyz}=3\)

vậy A=3