Question

Answer 1

`\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}=0`

`<=>\sqrt{x^2-4x+4}=\sqrt{x^2+4x+4}`

`<=>x^2-4x+4=x^2+4x+4`

`<=>8x=0`

`<=>x=0`

Vậy `S={0}`

Answer 2

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

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+2\right)^2}=0\)

\(ĐK:x\in R\)

\(\Leftrightarrow\left|x-2\right|-\left|x+2\right|=0\)

`@`Với \(x\ge2\)

`->` pt: \(x-2-x-2=0\)

           \(\Leftrightarrow-4=0\left(vô.lý\right)\)

`@`Với \(x< 2\)

`->`pt \(2-x-x-2=0\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy \(S=\left\{0\right\}\)