GTNN=2 <=> x=2

A=[(x-1)(x+6)][(x+2)(x+3)]

=(x²+5x-6)(x²+5x+6)

=(x²+5x)²-36

Ta thấy (x²+5x)²  >=0 nên (x²+5x)²-36 >=-36

Vậy GTNN của A là -36

dk \(1\le x\le3\)

\(P^2=x-1+3-x+2\sqrt{\left(x-1\right)\left(3-x\right)}\) =\(2+2\sqrt{\left(x-1\right)\left(3-x\right)}\)

ta co \(p^2\ge2\Rightarrow p\ge\sqrt{2}\) dau = xay ra khi \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

\(P^2=2+2\sqrt{\left(x-1\right)\left(3-x\right)}\le2+x-1+3-x=4\) (ap dung bdt amgm)\(\Rightarrow p\le2\)

dau = xay ra khi \(x-1=3-x\Leftrightarrow x=2\) 

kl min p= \(\sqrt{2}khi\orbr{\begin{cases}x=1\\x=3\end{cases}}\) maxp= 2 khix=2

\(\text{Đ}\text{ể}Pc\text{ó}ngh\text{ĩa}\Leftrightarrow\sqrt{x-1}\ge0\Leftrightarrow x-1\ge0\Leftrightarrow x\ge1\)>=1\(v\text{à}\sqrt{3-x}\ge0\Leftrightarrow3-x\ge0\Leftrightarrow x\le3\).\(x\ge1V\text{à}x\le3\Rightarrow PKh\text{ô}ngC\text{ó}Ngh\text{ĩa}\)

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)\)

\(=\left(x^2+3x+1\right)^2-1\ge-1\) với moi x

Dấu "=" xảy ra <=> x²+3x+1=0

<=>\(\left(x+\frac{3}{2}\right)^2-\frac{5}{4}=0< =>\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2=0\)

\(< =>\left(x+\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x+\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=0\)

<=>..... (x có 2 nghiệm)

Vậy Min của...=-1 khi.............