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Question Number 169768 by mnjuly1970 last updated on 08/May/22
[x]+[x+12]+[x−13]=8 [2x]=8−[x−13]=k k2⩽x<k+12,8−k+13⩽x<9−k+13 k2<28−3k3⇒9k<56 k<569★ 253−k<k+12⇒50−6k<3k+3 ⇒479<x★★ ★&★★:k=6 3⩽x<72(1) 73⩽x<103(2) (1)&(2)⇒73⩽x<103
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![[x]+[x+(1/2)]+[x−(1/3)]= 8 [2x]=8−[x−(1/3)]=k (k/2)≤x<((k+1)/2) , 8−k+(1/3)≤x <9−k+(1/3) (k/2) < ((28−3k)/3) ⇒ 9k < 56 k <((56)/9) ★ ((25)/3) −k<((k+1)/2) ⇒ 50−6k<3k+3 ⇒ ((47)/9) <x ★★ ★ & ★★ : k=6 3≤x< (7/2) (1) (7/3)≤x< ((10)/3) (2) (1) & (2) ⇒ (7/3)≤x<((10)/3)](Q169768.png)