\(A=\left(x+2\right)^2-5\ge-5\)
Dấu ''='' xảy ra <=> x = -2
Vậy GTNN A là -5 <=> x = -2
\(A=\left(x+2\right)^2-5\)
Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow\left(x+2\right)^2-5\ge-5\forall x\)
\(\Rightarrow A\ge-5\)
Dấu " = " xảy ra \(\Leftrightarrow x+2=0\)\(\Leftrightarrow x=-2\)
Vậy \(minA=-5\)\(\Leftrightarrow x=-2\)
\(\frac{s}{2\cdot v2}+\frac{s}{v2}=\frac{1}{4}\)
\(\frac{s}{v2}\left(\frac{1}{2}+1\right)=\frac{1}{4}\)
\(\frac{s}{v2}\cdot\frac{3}{2}=\frac{1}{4}\)
\(\frac{s}{v2}=\frac{1}{4}:\frac{3}{2}=\frac{1}{6}\)
với s=3,6
\(\Rightarrow\frac{18}{5}:v2=\frac{1}{6}\)
\(\Rightarrow v2=\frac{18}{5}:\frac{1}{6}=21,6\)