Question

50+2×21+4×[5+40:(19+3×a)]=116

Answer 1

50 + 2 × 21 + 4 × [ 5 + 40 : ( 19 + 3 × a ) ] = 116

50 + 42 + 4 × [ 5 + 40 : ( 19 + 3 × a ) ] = 116

42 + 4 × [ 5 + 40 : ( 19 + 3 × a ) ] = 116 - 50

42 + 4 × [ 5 + 40 : ( 19 + 3 × a ) ] = 66

4 × [ 5 + 40 : ( 19 + 3 × a ) ] = 66 - 42

4 × [ 5 + 40 : ( 19 + 3 × a ) ] = 24

[ 5 + 40 : ( 19 + 3 × a ) ] = 24 : 4

[ 5 + 40 : ( 19 + 3 × a ) ] = 6

[ 40 : ( 19 + 3 × a ) ] = 6 - 5

[ 40 : ( 19 + 3 × a ) ] = 1

19 + 3 × a = 1 × 40

19 + 3 × a = 40

3 × a = 40 - 19

3 × a = 21

a = 21 : 3

a = 7

Answer 2

Răspuns

7 ;

Explicație pas cu pas:

50+2x21+4x[5+40:(19+3a)] = 116 ;

42+4x[5+40:(19+3a)] = 116-50 = 66 ;

4x[5+40:(19+3a)] = 66-42 = 24 ;

5+40:(19+3a) = 24:4 = 6 ;

40:(19+3a) = 6-5 = 1 ;

19+3a = 40:1 = 40 ;

3a = 40-19  = 21 ;

a = 21:3 = 7