\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)=\left(3x+2\right)\left(3x-2\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x+2\right)\left(3x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow3x+2=0\) ; \(x+1=0\) ; \(-2x+1=0\)
+) \(3x+2=0\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
+) \(x+1=0\)
\(\Leftrightarrow x=-1\)
+) \(-2x+1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Tập nghiệm: \(S=\left\{\dfrac{-2}{3};\dfrac{1}{2};-1\right\}\)
(3x+2)(x2-1)=(9x2-4)(x+1)
<=> (3x+2)(x+1)(x-1)-(3x-2)(3x+2)(x+1)=0
<=> (3x+2)(x+1)(x-1-3x+2)=0
<=> (3x+2)(x+1)(1-2x)=0
<=> \(\left\{{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy....\(\left\{{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=-2\\x=-1\\-2x=-1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
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