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gaseste valoarea lui x care face adevarata fiecare relatie: b) [(x-416:4+2)x5+3]x2=1986 d) 5xx-2000:50-(1291-1091)=40

Răspuns :

b) [(x-416:4+2)×5+3] ×2=1986
[(x-104+2)×5+3]=1986:2
(x-102)×5=993-3
(x-102)×5=990
x-102=198
x=300

 d) 5×x-2000:50-(1291-1091)=40
5×x-40-200=40
5×x-240=40
5×x=40 +240
5×x=280
x=280:5
x=56