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Question Number 189509 by normans last updated on 18/Mar/23

Answered by mr W last updated on 18/Mar/23

Commented by mr W last updated on 18/Mar/23

(xk)^2 =(x−3)k×(2+x+3)k ⇒x=((15)/2) (2×1^2 +2^2 +x^2 +(2+x)^2 )k^2 =610 (6+((15^2 )/4)+(((19)/2))^2 )k^2 =610 ⇒k=2 R=(1+x)k =(1+((15)/2))×2=17

(xk)2=(x−3)k×(2+x+3)k⇒x=152(2×12+22+x2+(2+x)2)k2=610(6+1524+(192)2)k2=610⇒k=2R=(1+x)k=(1+152)×2=17

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