Theo bài ra , ta có :
\(\left(x-1\right)x\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt x2 + x = z =) x2 + x - 2 = z - 2
\(\Rightarrow z\left(z-2\right)=24\)
\(\Leftrightarrow z^2-2z=24\)
\(\Leftrightarrow z^2-2z-24=0\)
\(\Leftrightarrow\left(z+4\right)\left(z-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}z=-4\\z=6\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x^2+x=-4\\x^2+x=6\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-3\end{cases}}\)
Vậy S = { -1/2 ; -3 }
b)
\(x^4+3x^3+4x^2+3x+1=0\)
\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\left(x^2+x+1\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)(1)
Ta có :
\(x^2+x+1\)
\(\Leftrightarrow x^2+2\times\frac{1}{2}x+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\in Z\)(2)
Từ (1) và (2) suy ra phương trình có dạng
\(\left(x+1\right)^2=0\)( Vì phương trình (2) luôn lớn hơn 0 )
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy S = {-1}
Chúc bạn học tốt =))
Bài này có bắt tìm nghiệm \(x\in Z\)(2) đâu
