Question

Tìm x biết:

\(\sqrt{4x^2-4x+1}+\sqrt{2x-1}=0\)

Answer 1

`\sqrt{4x^2-4x+1}+\sqrt{2x-1}=0`          `ĐK: x >= 1/2`

`<=>\sqrt{(2x-1)^2}+\sqrt{2x-1}=0`

`<=>|2x-1|+\sqrt{2x-1}=0`

 Vì `x >= 1/2<=>2x-1 >= 0=>|2x-1|=2x-1`

  `=>2x-1+\sqrt{2x-1}=0`

`<=>\sqrt{2x-1}(\sqrt{2x-1}+1)=0`

   Vì `\sqrt{2x-1}+1 > 0`

`=>\sqrt{2x-1}=0`

`<=>2x-1=0`

`<=>x=1/2` (t/m)

Vậy `S={1/2}`

Answer 2

\(\Leftrightarrow\left|2x-1\right|=-\sqrt{2x-1}\)

=>2x-1=0

hay x=1/2

Answer 3

`=> sqrt ( 2x - 1)^2  = - sqrt (2x - 1)`

`=> | 2x - 1| = - sqrt ( 2x - 1)`

`=> 2x - 1 = 0`

`=> x = 1/2`.