Giải:
\(\sqrt{\dfrac{-3}{2+x}}=2\)
ĐK: \(\dfrac{-3}{2+x}\ge0\Leftrightarrow x< -2\)
\(\Leftrightarrow\left(\sqrt{\dfrac{-3}{2+x}}\right)^2=2^2\)
\(\Leftrightarrow\dfrac{-3}{2+x}=4\)
\(\Rightarrow-3=4\left(2+x\right)\)
\(\Leftrightarrow-3=8+4x\)
\(\Leftrightarrow-3-8=4x\)
\(\Leftrightarrow-11=4x\)
\(\Leftrightarrow x=\dfrac{-11}{4}\) (thỏa mãn)
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