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in Integrals calculus by (20 points)

If positive integers satisfying: (l+x)(l+y)=4y

   (l+y)(l+z)=4z

(l+z)(l+x)=4x then the value of 2x+3y+4z+x²+y²+z⁴ is 

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1 Answer

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by (44.6k points)

All equations (1 + x)(1 + y) = 4y

(1 + y) (1 + z) = 4z

(1 + z) (1 + x) = 4x

will satisfy by x = y = z = 1

\(\therefore\) 2x + 3y + 4z + x2 + y3+z4 

 = 2 + 3 + 4 + 1 + 1 + 1 = 12

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