\(C=-\left|x+\dfrac{5}{3}\right|\\ Ta.có:\left|x+\dfrac{5}{3}\right|\ge0\forall x\in R\\ \left|x+\dfrac{5}{3}\right|_{\left(Min\right)}=0\left(khi:x=-\dfrac{5}{3}\right)\\ Vậy:C_{max}=-\left|x+\dfrac{5}{3}\right|=0\left(khi:x=-\dfrac{5}{3}\right)\)
\(D=2-\left|3-x\right|\\ Vì:\left|3-x\right|\ge0\forall x\in R\\ \Rightarrow D=2-\left|3-x\right|\le2\forall x\in R\\ Vậy:D_{max}=2\left(khi:x=3\right)\)