Ta có: \(N=\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)\)
\(=\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+5\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x-10\right)\)
\(=\left(x^2+3x\right)^2-14\left(x^2+3x\right)+40\)
\(=\left(x^2+3x\right)^2-14\left(x^2+3x\right)+49-9\)
\(=\left(x^2+3x-7\right)^2-9\)
Ta có: \(x^2+3x-7\)
\(=x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\)
Ta có: \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\forall x\)
hay \(x^2+3x-7\ge-\frac{37}{4}\forall x\)
\(\Rightarrow\left(x^2+3x-7\right)^2\ge\frac{1369}{16}\forall x\)
\(\Rightarrow\left(x^2+3x-7\right)^2-9\ge\frac{1225}{16}\forall x\)
Dấu '=' xảy ra khi \(x+\frac{3}{2}=0\)
hay \(x=-\frac{3}{2}\)
Vậy: Giá trị nhỏ nhất của biểu thức \(N=\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)\) là \(\frac{1225}{16}\) khi \(x=-\frac{3}{2}\)
