a) We have |2x+3| + 2 \(\ge\)

When that, |2x+3| = 0 \(\Leftrightarrow\) 2x + 3 = 0 \(\Leftrightarrow\) x = - 1,5.

Hence, the smallest value of |2x+3| + 2 is 2 at x = -1,5.

b, We have 

(x+2)²\(\ge\) 0 \(\Rightarrow\)(x+2)²+ 4 \(\ge\) 4.

So, the smallest value of this expression is 4.

When that, (x+2)² = 0 \(\Leftrightarrow\) x + 2 = 0 \(\Leftrightarrow\) x = -2.

Hence, the smallest value of  x² + 4x + 8 is 4 at x = -2.

We have |x+2| = 2x -1 \(\Rightarrow\) 2x - 1 \(\ge\) 0 \(\Rightarrow\) x \(\ge\)\(\dfrac{1}{2}\)

1\(^{st}\)case: x \(\ge\) 0 \(\Rightarrow\)|x+2| = x+2 \(\Rightarrow\)x + 2 = 2x -1 \(\Rightarrow\)  x = 3 (satisfied)

2\(^{nd}\)case: x < 0 \(\Rightarrow\) |x+2| = - (x+2) \(\Rightarrow\)-x -2 = 2x - 1\(\Rightarrow\)x = \(\dfrac{-1}{3}\)(absurd)

Hence, x = 3.