Question

tìm x

Answer 1

`\sqrt{x^2+6x+9}=5-2x`        `ĐK: x <= 5/2`

`<=>\sqrt{(x+3)^2}=5-2x`

`<=>|x+3|=5-2x`

`<=>[(x+3=5-2x),(x+3=2x-5):}`

`<=>[(x=2/3 (t//m)),(x=8(ko t//m)):}`

Vậy `S={2/3}`

Answer 2

\(\sqrt{x^2+6x+9}=5-2x\)

\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=5-2x\)

\(\Leftrightarrow\left|x+3\right|=5-2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=5-2x\\x+3=-5+2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=2\\-x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=8\end{matrix}\right.\)