Ajutati ma la ex 529. Punctele c si d.
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c)406-4×[a×(30-20÷5÷4)]=58
406-4×[a×(30-1)]=58
406-4×(a×29)=58
406-4×29a=58
4×29a=406-58
4×29a=348
29a=348÷4
29a=87
a=87÷29=3
c)406-4x[ax(30-20:5:4)]=58 |+406
4x(ax29)= 464 |:4
ax29= 116 |:29
a=4
d)[21-(648:24+a):7]x3+55=100|-55
[21-(162+a):7]x3=45 |:3
21-(162+a):7=15 | +21
(162+a):7= 36 |x7
162+a= 252 |-162
a= 90