\(A=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)

\(Min_A=1\)

\(B=\left|x+1\right|+\left|x-1\right|\ge\left|x+1-x+1\right|=2\)

\(Min_B=2\)

\(C=\left|2x+3\right|+\left(y-1\right)^2-5\ge-5\)

\(Min_C=-5\)

\(E=\left(2x+3\right)^2-4\ge-4\)

\(Min_E=-4\)

\(n\left(n+3\right)=n^2+3n\)

\(\left(n+2\right)\left(n+1\right)=n^2+3n+2\)

Vì \(n^2+3n< n^2+3n+2\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\left(n\in N\right)\)

b) \(\dfrac{n}{2n+1}=\dfrac{3n}{6n+3}< \dfrac{3n+1}{6n+3}\)

c) \(\dfrac{10^8+2}{10^8-1}=1+\dfrac{1}{10^8-1}\)

\(\dfrac{10^8}{10^8-3}=\left(1+\dfrac{3}{10^8-3}\right)\)

Vì \(\dfrac{1}{10^8-1}>\dfrac{3}{10^8-3}\Rightarrow\dfrac{10^8+2}{10^8-1}< \dfrac{10^8}{10^8-3}\)