`#3107`
a)
`8x^2 - 5 = 67 `
`=> 8x^2 = 67 + 5`
`=> 8x^2 = 72`
`=> x^2 = 9`
`=> x^2 = (+-3)^2`
`=> x = +-3`
Vậy, `x \in {-3;3}`
b)
`(4x - 2)^4=16`
`=> (4x - 2) = (+-2)^4`
`=> ` TH1: `4x - 2 = 2`
`=> 4x = 4`
`=> x =1`
TH2: `4x - 2 = -2`
`=> 4x = 0`
`=> x=0`
Vậy, `x \in {0; 1}.`
a) \(8x^2-5=67\)
\(8x^2=72\)
\(x^2=9\)
\(x=3\)
Vậy x = 3
b) \(\left(4x-2\right)^4=16\)
\(\left(4x-2\right)^2=4\)
\(4x-2=2\)
\(4x=4\)
\(x=1\)
Vậy x = 1