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Question Number 99218 by 9027201563 last updated on 19/Jun/20

Commented by 9027201563 last updated on 19/Jun/20

$p l s k n e e d s o l u t i o n i m m e d i a t e l y$
Commented by bobhans last updated on 19/Jun/20
$(1)+(2)+(3) ⇒x+y+z+(9/x)+(4/y)+((16)/z) = 18 (x+(9/x))+(y+(4/y))+(z+((16)/z)) = 18 (x+(9/x)−6)+(y+(4/y)−4)+(z+((16)/z)−8) = 0 { ((x^2 −6x+9 = 0 ⇒x = 3)),((y^2 −4y+4 = 0 ⇒y= 2)),((z^2 −8z+16 = 0 ⇒z = 4 )) :}$
$(1) + (2) + (3) \Rightarrow x + y + z + \frac{9}{x} + \frac{4}{y} + \frac{16}{z} = 18$ $(x + \frac{9}{x}) + (y + \frac{4}{y}) + (z + \frac{16}{z}) = 18$ $(x + \frac{9}{x} - 6) + (y + \frac{4}{y} - 4) + (z + \frac{16}{z} - 8) = 0$ ${\begin{cases} x^{2} - 6 x + 9 = 0 \Rightarrow x = 3 \\ y^{2} - 4 y + 4 = 0 \Rightarrow y = 2 \\ z^{2} - 8 z + 16 = 0 \Rightarrow z = 4 \end{cases}$
Commented by 9027201563 last updated on 19/Jun/20
Thanks Alot I really appreciate
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