a) Đặt \(x^2+4x+3=0\)
\(\Leftrightarrow x^2+x+3x+3=0\)
\(\Leftrightarrow x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy: S={-1;-3}
b) Ta có: \(x^2+x+2\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\forall x\)
hay \(x^2+x+2>0\forall x\)
Vậy: \(S=\varnothing\)
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