4.(3x + 1) = 7.(x + 2)
12x + 4 = 7x + 14
12x - 7x = 14 - 4
5x = 10
x = 10 : 5
x = 2
Vậy x = 2
\(\dfrac{4}{x+2}=\dfrac{7}{3x+1}\)
\(=>4\left(3x+1\right)=\left(x+2\right)7\)
\(=>12x+4=7x+14\)
\(=>12x-7x=14-4\)
\(=>5x=10\)
\(=>x=2\)
`4/(x+2) = 7/(3x+1)`
Đk: `{(x + 2 ne 0), (3x + 1 ne 0):} <=> {(x ne -2), (x ne -1/3):}`
`-> 7x + 14 = 12x + 4`
`-> 10 = 5x`
`-> x = 2 (tm)`