Let x,y, and z be real numbers such that x + y + z = 6 and \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) = 2. Find \(\frac{x + y}{z} + \frac{y + z}{x} + \frac{x + z}{y}.\)
+6248 \(\dfrac 1 x + \dfrac 1 y + \dfrac 1 z = 2\\ \dfrac{x+y+z}{x+y+z}\left( \dfrac 1 x + \dfrac 1 y + \dfrac 1 z\right) = 2\\ \dfrac 1 6 \left(1+\dfrac{y+z}{x} + 1 + \dfrac{x+z}{y}+1+\dfrac{x+y}{z}\right) = 2\\ 3 + \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 12\\ \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 9\)
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+128475 x + y + z = 6 (1)
1/x + 1/y + 1/z = 2 ⇒ [ yz + xz + xy] / xyz = 2 ⇒ [ xy + xz + yz] = 2xyz (2)
x + y y + z x + z
Find ____ + ______ + ______ =
z x y
xy ( x + y) + yz( y + z) + xz ( x + z)
________________________________ =
xyz
xy ( 6 - z) + yz ( 6 - x) + xz ( 6 -y)
______________________________ =
xyz
6xy - xyz + 6yz - xyz + 6xz - xyz
_______________________________ =
xyz
6 [ xy + xz + yz ] - 3 [ xxz ]
_______________________ =
xyz
6 [ 2xyz ] - 3 [xyz]
________________ =
xyz
xyz [ 6*2 - 3 ]
____________ =
xyz
6*2 - 3 =
12 - 3 =
9
