Question

\(\sqrt{x+2\sqrt{2x-4}}=5\) tìm x

 

Answer 1

`\sqrt{x+2\sqrt{2x-4}}=5`          `ĐK: x >= 2`

`<=>x+2\sqrt{2x-4}=25`

`<=>2\sqrt{2x-4}=25-x`   `(2 <= x <= 25)`

`<=>8x-16=625-50x+x^2`

`<=>x^2-58x+641=0`

`<=>x^2-2x.29+841-200=0`

`<=>(x-29)^2=200`

`<=>|x-29|=10\sqrt{2}`

`<=>x=29+-10\sqrt{2}` 

  Mà `2 <= x <= 25`

 `=>x=29-10\sqrt{2}`

Answer 2

ĐKXĐ \(x\ge2\)

\(\sqrt{x+2\sqrt{2x-4}}=5\)

\(\Leftrightarrow\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}=5\)

\(\Leftrightarrow\left|\sqrt{x-2}+\sqrt{2}\right|=5\)

\(\Leftrightarrow\sqrt{x-2}+\sqrt{2}=5\)

\(\Leftrightarrow\sqrt{x-2}=5-\sqrt{2}\)

\(\Rightarrow x=29-10\sqrt{2}\)