a: \(3x^2+4x+4=0\)
\(\Delta=4^2-4\cdot3\cdot4=16-48=-32< 0\)
=>Phương trình vô nghiệm
b: \(2x^2-x-6=0\)
=>\(2x^2-4x+3x-6=0\)
=>2x(x-2)+3(x-2)=0
=>(x-2)(2x+3)=0
=>\(\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(3x^2+4x+4=0\)
\(\Leftrightarrow3\left(x^2+\dfrac{4}{3}x\right)+4=0\)
\(\Leftrightarrow3\left[x^2+2\cdot x\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2\right]-3\cdot\dfrac{4}{9}+4=0\)
\(\Leftrightarrow3\left(x+\dfrac{2}{3}\right)^2+\dfrac{8}{3}=0\)
\(\Rightarrow x\in\varnothing\) (vì \(3\left(x+\dfrac{2}{3}\right)^2+\dfrac{8}{3}>0\forall x\))
$---$
\(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
#$\mathtt{Toru}$
\(3x^2+4x+4=0\)
Mà: \(3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{3}\right)=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{8}{9}\right)=3\left(x+\dfrac{2}{3}\right)^2+\dfrac{8}{3}>0\)
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\(2x^2-x-6=0\\ \Leftrightarrow2x^2-4x+3x-6=0\\ \Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
