\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)-3=0\)
\(\Rightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-3=0\)
\(\Rightarrow\left(x^2+x\right)\left(x^2+2x-x-2\right)-3=0\)
\(\Rightarrow\left(x^2+x\right)\left(x^2+x-2\right)-3=0\)
Đặt \(t=x^2+x\) thì \(x^2+x-2=t-2\)
Ta có biểu thức theo t:
\(t\left(t-2\right)-3=0\)
\(\Rightarrow t^2-2t-3=0\)
\(\Rightarrow t^2-3t+t-3=0\)
\(\Rightarrow t\left(t-3\right)+\left(t-3\right)=0\)
\(\Rightarrow\left(t-3\right)\left(t+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t-3=0\\t+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=3\\t=-1\end{matrix}\right.\)
Thay \(t=x^2+x\) ta có:
\(\Rightarrow\left[{}\begin{matrix}x\left(x+1\right)=3\\x\left(x+1\right)=-1\end{matrix}\right.\)
Phần sau bạn tự làm nhé!
Chúc bạn học tốt!
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)-3=0\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-3=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-3=0\)
Đặt \(x^2+x=t\) , ta có :
\(t\left(t-2\right)-3=0\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left(t^2-2t+1\right)-4=0\)
\(\Leftrightarrow\left(t-1\right)^2-2^2=0\)
\(\Leftrightarrow\left(t-1-2\right)\left(t-1+2\right)=0\)
\(\Leftrightarrow\left(t-3\right)\left(t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-3=0\\t+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-1\end{matrix}\right.\)
Với t = 3
\(\Rightarrow x^2+x=3\)
\(\Leftrightarrow x^2+x-3=0\left(1\right)\)
Ta có :
\(x^2+x-3\)
\(\Leftrightarrow\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{11}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)
=> pt (1) vô nghiệm
Với t = -1
\(\Leftrightarrow x^2+x=-1\)
\(\Leftrightarrow x^2+x+1=0\) ( vô lý ) c/m tương tự trên
Vậy pt vô nghiệm
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-3=0\) (1)
Đặt \(x^2+x=a\)
(1) \(\Leftrightarrow a\left(a-2\right)-3=0\)
\(\Leftrightarrow a^2-2a-3=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left(x^2+x-3\right)\left(x^2+x+1\right)=0\)
