Question

giải pt

\(\sqrt{x^2+x}=x\)

Answer 1

Ý bạn là \(\sqrt{x^2+x}=x\) hay \(\sqrt{x^2}+x=x?\)

Answer 2

\(\sqrt{x^2+x}=x\)

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

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

`<=> x = 0`

vậy `S = {0}`

Answer 3

\(\sqrt{x^2+x}=x\)

\(x^2+x=x^2\)

\(x=0\)

Vậy x= 0

Answer 4

ĐKXĐ : \(x^2+x\ge0\Leftrightarrow x\left(x+1\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x+1\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge-1\end{matrix}\right.\Rightarrow x\ge0\\\left\{{}\begin{matrix}x\le0\\x\le-1\end{matrix}\right.\Rightarrow x\le-1\end{matrix}\right.\)

 

Ta có : \(\sqrt{x^2+x}=x\)

\(\Leftrightarrow x^2+x=x^2\Leftrightarrow x=0\left(tmđk\right)\)

Vậy : Phương trình có tập nghiệm \(S=\left\{0\right\}\)

Answer 5