Question

a,\(\left(x-\frac{1}{2}\right)^2=0\)                        

b, \(\left(x-2\right)^2=1\)  

c, \(\left(2x-1\right)^3=-8\)

d, \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

Answer 1

a, x= \(\frac{1}{2}\)

b, \(x=1;x=3\)

c,\(x=1,5\)

d, x= \(\frac{-1}{4}\)

Answer 2

bạn làm chi tiết hơn đc k 

Answer 3

a; ( x - \(\frac{1}{2}\))²  =0
< = > x - \(\frac{1}{2}\) = 0
< = > x = \(\frac{1}{2}\)

b; ( x - 2)² = 1
< = > x - 2 = 1 hoặc x - 2 = -1

Với x - 2 = 1 => x = 3
Với x - 2 = -1 => x = 1 

c; ( 2x - 1 )\(^3\) = -8
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(\frac{-1}{2}\)

d; ( x + \(\frac{1}{2}\))^{\(^2\)} = \(\frac{1}{16}\)

<=> x + \(\frac{1}{2}\)= \(\frac{1}{4}\)hoặc x +\(\frac{1}{2}\)= - \(\frac{1}{4}\)

Với x + \(\frac{1}{2}\)= \(\frac{1}{4}\)=> x = - \(\frac{1}{4}\)

Với x + \(\frac{1}{2}\)= - \(\frac{1}{4}\)=> x = - \(\frac{3}{4}\)