Ta có: \(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-32\)
\(\Leftrightarrow A=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-32\)
\(\Leftrightarrow A=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-32\)
Đặt \(a=x^2+5x+4\) \(\Rightarrow\)\(a+2=x^2+5x+6\)
Ta lại có: \(A=a.\left(a+2\right)-32\)
\(\Leftrightarrow A=a^2+2a-32\)
\(\Leftrightarrow A=a^2+2a-32\)
\(\Leftrightarrow A=\left(a^2+2a+1\right)-33\)
\(\Leftrightarrow A=\left(a+1\right)^2-\left(\sqrt{33}\right)^2\)
\(\Leftrightarrow A=\left(a+1-\sqrt{33}\right)\left(a+1+\sqrt{33}\right)\)
\(\Leftrightarrow A=\left(x^2+5x+5-\sqrt{33}\right)\left(x^2+5x+5+\sqrt{33}\right)\)
Ta có A = (x + 1)(x + 2)(x + 3)(x + 4) - 32
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 32
= (x2 + 5x + 4)(x2 + 5x + 6) - 32
= (x2 + 5x + 5 - 1)(x2 + 5x + 5 + 1) - 32
= (x2 + 5x + 5)2 - 1 - 32
= (x2 + 5x + 5)2 - 33 \(\ge-33\)
Dấu "=" xảy ra <=> x2 + 5x + 5 = 0
=> (x2 + 5x + 25/4) = 5/4
=> (x + 5/2)2 = 5/4
=> \(\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)
Vậy Min A = -33 <=> \(x\in\left\{\sqrt{\frac{5}{4}}-\frac{5}{2};-\sqrt{\frac{5}{4}}-\frac{5}{2}\right\}\)
