a) \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-1=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-1=-5\\x^2+x-1=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}=-5\\\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=-\dfrac{15}{4}\left(VL\right)\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{25}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{2;-3\right\}\)
b) \(2x^3+9x^2+7x-6=0\)
\(\Leftrightarrow2x^3-x^2+10x^2-5x+12x-6=0\)
\(\Leftrightarrow x^2\left(2x-1\right)+5x\left(2x-1\right)+6\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[\left(x^2+2x\right)+\left(3x+6\right)\right]\left(2x-1\right)=0\)
\(\Leftrightarrow\left[x\left(x+2\right)+3\left(x+2\right)\right]\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình đã cho là \(S=\left\{-3;-2;\dfrac{1}{2}\right\}\)
a) x(x+1)(x-1)(x+2) =24
<=> (x^2+x)(x^2+x-2)=24
Đặt x^2+x=t
=> t(t-2)=24
<=> t^2-2t=24
<=>t^2-2t-24=0
<=> (t+4)(t-6)=0
<=> t=-4=>x^2+x=-4<=>x^2+x+4=0<=> x=-0,5
hoặc t=6=> x^2+x=6<=> x=2 hoặc x=-3
