Question

Chờ x,y,z khác 0 và \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

Chứng minh rằng : \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)

MN giúp mk với !

Answer 1

Ta có: \(\left(x+y\right)+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=x^2+y^2+z^2\)

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

Hay \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{-1}{z}\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)

\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{3}{xy}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)hay \(\dfrac{1}{x^3}-\dfrac{3}{xyz}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)

\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)