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y-2-1-x-2-4-z-2-9-10-x-y-z-




Question Number 142138 by mathdanisur last updated on 26/May/21
(√(y^2 +1)) + (√(x^2 +4)) + (√(z^2 +9)) = 10  x+y+z=?
y2+1+x2+4+z2+9=10x+y+z=?
Commented by mathdanisur last updated on 27/May/21
cool sir thank you
coolsirthankyou
Commented by mr W last updated on 27/May/21
(√(y^2 +1^2 ))+(√(x^2 +2^2 ))+(√(z^2 +3^2 ))≥(√((x+y+z)^2 +(1+2+3)^2 ))  ⇒(x+y+z)^2 ≤10^2 −6^2 =64=8^2   ⇒−8≤x+y+z≤8
y2+12+x2+22+z2+32(x+y+z)2+(1+2+3)2(x+y+z)210262=64=828x+y+z8
Commented by MJS_new last updated on 26/May/21
no unique value
nouniquevalue
Commented by MJS_new last updated on 26/May/21
(√(x^2 +4))=a ⇔ x=±(√(a^2 −4)); 2≤a≤10  (√(y^2 +1))=b⇔ y=±(√(b^2 −1)); 1≤b≤10  (√(z^2 +9))=c⇔ z=±(√(c^2 −9)); 3≤c≤10  a+b+c=10  infinite values for  x+y+z=±(√(a^2 −4))±(√(b^2 −1))±(√(c^2 −9))
x2+4=ax=±a24;2a10y2+1=by=±b21;1b10z2+9=cz=±c29;3c10a+b+c=10infinitevaluesforx+y+z=±a24±b21±c29
Commented by MJS_new last updated on 26/May/21
−8≤x+y+z≤8
8x+y+z8
Commented by mathdanisur last updated on 29/May/21
cool Sir thanks
coolSirthanks

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