Question

9.
1 1 1 5
If x + y + z = 19 xyz = 126.-+-+--
X Y Z
and x > 0, y > 0, z>0, then the value of
x² + y² + z² is
(a) 161
(c) 181
(b) 171
(5) 191​

Answer 1

Step-by-step explanation:

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