Sa se calculeza 3a -12,daca a verifica egalitatea 10-10÷{1+3×[(39-60÷a×2)+21]÷10=9
Răspunsuri la întrebare
Salut!
10-10÷{1+3×[(39-60÷a×2)+21]÷10}=9
10÷{1+3×[(39-60÷a×2)+21]÷10}=10-9
1+3×[(39-60÷a×2)+21]÷10=1x10
1+3×[(39-60÷a×2)+21]=10x10
(39-60÷a×2)+21=(100-1):3
39-60:ax2=30-21
39-60:ax2=9
60:ax2=39-9
60:a=30:2
a=60:15
a=4
3a-12=
3x4-12=
12-12=
=0
Succes!
Răspuns:
Explicație pas cu pas:
10 - 10 : { 1 + 3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 } = 9
10 : { 1 + 3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 } = 10 - 9
10 : { 1 + 3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 } = 1
1 + 3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 = 10 : 1
1 + 3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 = 10
3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 = 10 - 1
3 × [ ( 39 - 60 : a × 2 ) + 21] : 10 = 9
( 39 - 60 : a × 2 ) + 21 = 9 × 10 : 3 → aplic metoda mersului invers
( 39 - 60 : a × 2 ) + 21 = 90 : 3
( 39 - 60 : a × 2 ) + 21 = 30
39 - 60 : a × 2 = 30 - 21
39 - 60 : a × 2 = 9
60 : a × 2 = 39 - 9
60 : a × 2 = 30
60 : a = 30 : 2
60 : a = 15
a = 60 : 15
a = 4
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3 × a - 12 = 3 × 4 - 12 = 12 - 12 = 0