Question

1) Află numerele necunoscute. (Prin metoda mersului invers).
a) (ax2 - 40):23=12
b) [(a-416: 4+2) ×5+3] × 2 = 1986
c) [(a−1): 100 +90] : 10 - 10 = 0
d) a: 2 + (123x3-53×4) = 257
e) 54-4x3: [2×11-5-(5×4-a)] + 14 × 2 = 76
f) 9× [3+4x (15 + 6x a): 18] = 81
g) [(a: 3-1:6 + 489]: 25+2=26
h) a×4+ (6×7-5×6-72:8) × 76 = 332
i) (651-72×4-306: 2): a x 9 = 45
j) 100x (6×9 - a x 7 - 4) = 100​

Answer 1

Răspuns:

a) (a×2-40)=12×23

(a×2-40)=276

a×2=276+40

a×2=316

a=316:2

a=157

b) (a-416:4+2)×5+3=1986:2

(a-416:4+2)×5+3=993

(a-416:4+2)×5=993-3

(a-416:4+2)×5=990

a-416:4+2=990:5

a-416:4+2=198

a-416:4=198-2

a-416:4=196

a-416=196:4

a-416=49

a=49+416

a=465

c) [(a-1):100+90]:10=0+10

[(a-1):100+90]:10=10

(a-1):100+90=10×10

(a-1):100+90=100

(a-1):100=100-90

(a-1):100=10

a-1=10×100

a-1=1000

a=1000+1

a=1001

d) a:2+(369-208)=257

a:2+101=257

a:2=257-101

a:2=146

a=146×2

a=292

e,f,g,h,i,j)nu stiu am incercat