Question

giải phương trình:  \( x+{ \sqrt{2x+3} }=0\)

giúp mình với

Answer 1

ĐKXĐ: \(2x+3\ge0\Leftrightarrow x\ge-\frac{3}{2}\\ \)

Ta có: \(x+\sqrt{2x+3}=0\Leftrightarrow\sqrt{2x+3}=-x\)

\(\Leftrightarrow\hept{\begin{cases}-x\ge0\\\left(\sqrt{2x+3}\right)^2=\left(-x\right)^2\end{cases}\Leftrightarrow\hept{\begin{cases}x\le0\\2x+3=x^2\end{cases}\Leftrightarrow}\hept{\begin{cases}-\frac{3}{2}\le x\le0\\x^2-2x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}-\frac{3}{2}\le x\le0\\\left(x-3\right).\left(x+1\right)=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le0\\\orbr{\begin{cases}x=3\\x=-1\end{cases}}\end{cases}\Leftrightarrow x=-1}\)\(\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le0\\x=3,x=-1\end{cases}\Leftrightarrow x=-1}\)

Vậy x=-1

\(\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le0\\\orbr{\begin{cases}x=3\\x=-1\end{cases}}\end{cases}\Leftrightarrow x=-1}\)

Answer 2

cảm ơn bạn nhiều nhiều ^-^

Answer 3

\(\sqrt{2x+3}=-x.\)

\(2x+3=x^2\)

\(x^2-2x-3=0\)\(\)

\(\left(x+1\right)\left(x-3\right)=0\)