Question

Answer 1

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

\(\Leftrightarrow\left(\sqrt{x-1}+1\right)^2=4\)

\(\Leftrightarrow\sqrt{x-1}+1=2\)

=>x-1=1

hay x=2

Answer 2

`\sqrt{x+2\sqrt{x-1}}=2`          `ĐK: x >= 1`

`<=>x+2\sqrt{x-1}=4`

`<=>x-1+2\sqrt{x-1}+1=4`

`<=>(\sqrt{x-1}+1)^2=4`

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

`@TH1:\sqrt{x-1}+1=2`

     `<=>\sqrt{x-1}=1`

    `<=>x-1=1<=>x=2` (t/m)

`@TH2:\sqrt{x-1}+1=-2`

    `<=>\sqrt{x-1}=-3` (Vô lí)

Vậy `S={2}`