Question

tim GTNN : M=(x+1)(x+2)(x+3)(x+4)

Answer 1

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

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

\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)

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

Đặt \(x^2+5x=t\) ,ta có :

\(\left(t+4\right)\left(t+6\right)\)

\(=t^2+4t+6t+24\)

\(=t^2+10t+24\)

\(=t^2+2.t.5+5^2-1\)

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

Ta có :

\(\left(t+5\right)^2\ge0\) với mọi x

\(\Rightarrow\left(t+5\right)^2-1\ge-1\) với mọi x

Dấu = xảy ra khi \(\left(t+5\right)^2=0\Rightarrow t+5=0\Rightarrow t=-5\)

Vậy \(Min_M=-1\Leftrightarrow x=-5\)