This made me won

Algebra Level 2

Let xx, yy, and zz be real numbers such that x+y+z=6 x+y+z=6 and 1x+1y+1z=2\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2 . Find the value of (x+y)z+(y+z)x+(x+z)y\dfrac{(x+y)}{z}+\dfrac{(y+z)}{x}+\dfrac{(x+z)}{y}.

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