Question

cho x,y, z khác 0 thỏa mãn \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=0.\)tính \(\dfrac{x^2}{yz}+\dfrac{z^2}{xy}+\dfrac{y^2}{xz}\)

Answer 1

Do \(xyz\ne0\) ta có:

\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)

Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)

\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)

Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)

\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)