\(\sqrt{x^2+x+3}=\frac{\sqrt{4\left(x^2+x+3\right)}}{2}=\frac{\sqrt{\left(2x+1\right)^2+11}}{2}\in Q\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2+11}\in Q\Leftrightarrow\left(2x+1\right)^2+11=y^2\text{ }\left(y\in N\right)\)
\(\Leftrightarrow\left(2x+1\right)^2-y^2=-11\)
\(\Leftrightarrow\left(2x+1-y\right)\left(2x+1+y\right)=-1.11=-11.1\)
\(\Rightarrow\hept{\begin{cases}2x+1-y=-11\\2x+1+y=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=6\end{cases}}}\)
hoặc \(\hept{\begin{cases}2x+1-y=-1\\2x+1+y=11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\end{cases}}\)
\(KL:x\in\left\{-3;2\right\}\)
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giải
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