\(T=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x+4=a\) \(\Rightarrow T=\left(a-1\right)\left(a+1\right)\)
\(=a^2-1=\left(x^2+5x+5\right)^2-1\ge-1\)
Vậy \(Min_T=-1\) khi
\(x^2+5x+5=0\Rightarrow\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{5}{4}=0\)\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=\dfrac{5}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x+\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}-\dfrac{5}{2}\\x=-\sqrt{\dfrac{5}{4}}-\dfrac{5}{2}\end{matrix}\right.\)
Câu hỏi của Tran Ngoc Hoang Khanh - Toán lớp 0 | Học trực tuyến
