Question

\(\sqrt{25x-50}-\dfrac{\sqrt{x-2}-1}{2}=8\sqrt{\dfrac{9x-18}{16}}\)

Answer 1

`\sqrt{25x-50}-[\sqrt{x-2}-1]/2=8\sqrt{[9x-18]/16}`        `ĐK: x >= 2`

`<=>\sqrt{5^2(x-2)}-[\sqrt{x-2}-1]/2=8\sqrt{[3^2(x-2)]/[4^2]}`

`<=>5\sqrt{x-2}-[\sqrt{x-2}-1]/2=6\sqrt{x-2}`

`<=>10\sqrt{x-2}-\sqrt{x-2}+1=12\sqrt{x-2}`

`<=>3\sqrt{x-2}=1`

`<=>\sqrt{x-2}=1/3`

`<=>x-2=1/9<=>x=19/9` (t/m)

Vậy `S={19/9}`

Answer 2

`\sqrt{25x-50}-(\sqrt{x-2}-1)/2=8\sqrt{(9x-18)/16}ĐKXĐ:x>=2`

`<=>\sqrt{25}\sqrt{x-2}+(1-\sqrt{x-2})/2=8\sqrt{9/16}\sqrt{x-2}`

`<=>5\sqrt{x-2}+(1-\sqrt{x-2})/2=6\sqrt{x-2}`

`<=>10\sqrt{x-2}+1-\sqrt{x-2}-12\sqrt{x-2}=0`

`<=>-3\sqrt{x-2}=-1`

`<=>\sqrt{x-2}=1/3`

`<=>x-2=1/9`

`<=>x=1/9+2`

`<=>x=19/9(n)`

Vậy `S={19/9}`