Question

\(6+\sqrt{x}-x=0\)

Answer 1

ĐK: `x>=0`

`6+\sqrtx-x=0`

`6+3\sqrtx-2\sqrtx-x=0`

`<=>3(2+\sqrtx)-\sqrtx(2+\sqrtx)=0`

`<=>(\sqrtx+2)(3-\sqrtx)=0`

`<=> [(\sqrtx+2=0),(3-\sqrtx=0):}`

`<=> [(\sqrtx=-2 (L)),(\sqrtx=3):}`

`<=>x=9`

Vậy `S={9}`.

Answer 2

ĐK: $x\ge 0$

$6+\sqrt{x}-x=0$

$6+3\sqrt{x}-2\sqrt{x}-x=0$

$\iff 3(2+\sqrt{x})-\sqrt{x}(2+\sqrt{x})=0$

$\iff (\sqrt{x}+2)(3-\sqrt{x})=0$

$\iff [\begin{array}{c}\sqrt{x}+2=0\\ 3-\sqrt{x}=0\end{array}$

$\iff [\begin{array}{c}\sqrt{x}=-2\left(L\right)\\ \end{array}$

Answer 3

Answer 4

\(\rightarrow x-\sqrt{x}-6=0\)

\(\rightarrow x-3\sqrt{x}+2\sqrt{x}-6=0\)

\(\rightarrow\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)=0\)

\(\rightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)

\(\rightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}+2=0\end{matrix}\right.\)

\(\rightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-2\left(voli\right)\end{matrix}\right.\)

\(\rightarrow x=9\)

Answer 5

Ta có: \(6+\sqrt{x}-x=0\)

\(\Leftrightarrow\left(3-\sqrt{x}\right)\left(2+\sqrt{x}\right)=0\)

\(\Leftrightarrow3-\sqrt{x}=0\)

hay x=9