Ta có :

\(M=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=\left[x^2+5x+4\right]\left[x^2+5x+6\right]\)

\(=\left[x^2+5x+5-1\right]\left[x^2+5x+5+1\right]\)

\(=\left(x^2+5x+5\right)^2-1\)

\(\Rightarrow Min_M=-1\Leftrightarrow x^2+5x+5=0\)

\(\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{5}{4}=0\)

\(\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)

\(\orbr{\begin{cases}x+\frac{5}{2}=\frac{\sqrt{5}}{2}\\x+\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{-5+\sqrt{5}}{2}\\x=-\frac{5+\sqrt{5}}{2}\end{cases}}\)

Vậy ...

\(=\left(-\frac{\sqrt{5}}{2}-\frac{3}{2}\right)\left(-\frac{\sqrt{5}}{2}-\frac{1}{2}\right)\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\)

   x^2 + 4y^2 - 2x + 10+ 4xy - 4y

= (x^2 + 4xy + 4y^2) - (2x + 4y) + 10

= (x + 2y)^2 - 2 (x + 2y) + 10

Thay x + 2y = 5 vào biểu thức trên ,ta  được :

   5^2 - 2 . 5 + 10

= 25 - 10 + 10

= 25

Giá trị nhỏ nhất của A=2018

Ta có : \(A=x^2+4x+2y^2+2xy+2018\)

\(\RightarrowđểAmin\)thì \(x^2+4x+2y^2+2xy=0\)

\(\Rightarrow Amin=0+2018=2018\)

\(\Rightarrow Amin=2018\)

   x² - 24x - 25

= x² + x - 25x - 25

= x(x + 1) - 25(x + 1)

= (x + 1) (x - 25)