`@` `\text {Ans}`
`\downarrow`
`b)`
\([(4x+28).3+55] \div 5=35\)
`=> (4x + 28) * 3 + 55 = 35 * 5`
`=> (4x + 28)*3 + 55 = 175`
`=> (4x + 28)*3 = 175 - 55`
`=> (4x + 28)*3 =120`
`=> 4x + 28 = 120 \div 3`
`=> 4x + 28 = 40`
`=> 4x = 40 - 28`
`=> 4x = 12`
`=> x = 12 \div 4`
`=>x = 3`
Vậy, `x = 3.`
=>[12x+84+55]=175
=>12x=175-84-55=120-84=36
=>x=36/12=3
<=>[12x+84+55]:5=35
<=>[12x+139]:5=35
<=>12x+139=35.5
<=>12x=175-139
<=>x=3