Question

[936-(8x100+a)]:[10+(35:5+8)]=4
[(72:a-3x4)x36-19x 5]-(124x6-301x2)=195 Cu demonstrare!!☺

Answer 1

a ese egal cu 3 
|(72/a-3*4)*36-19*5|-(124*6-301*2)=195
|(72/a-12)*36-95|-744-602=195
|(72/a-12)*36-95|-142=195
|(72/a-12)*36-95Z=195+142
|(72/a-12)*36-95|=337
|(72/a-12)*36=337+95
(72/a-12)*36=432
72/a-12=432/36
72/a-12=12
72/a=12+12
72/a=24
a=72/24
a=3

|936-(8*100+a)|/|10+(35/5+8)|=4
|936-(800+a)|/|(10+15)|=4
|936-(800+a)/25=4
936-(800+a)=4*25
936-(800+a)=100
800+a=936-100
800+a=836
a=836-800
a=36