`a)` Ta có: `(2x+3)^3≥0; (y-1/2)^2≥0`
`=> A= (2x+3)^2+(y-1/2)^2+2017≥0+0+2017`
`=> A≥2017`
`A_min``= 2017 <=>` $\begin{cases} (2x+3)^2= 0\\(y-1/2)^2= 0 \end{cases}$
`<=>` $\begin{cases} 2x+3)= 0\\y-1/2= 0 \end{cases}$
`<=>` $\begin{cases} x= -3/2\\y=1/2 \end{cases}$
`b)` Ta có: `2(x+1)^2≥0; |-3(x^2-1)|≥0`
`=> B= 2(x+1)^2+|-3(x^2+1)|≥0`
`B_min``= 0 <=>` $\begin{cases} x+1=0\\x²-1\end{cases}$
`<=>` $\begin{cases} x+1= 0\\(x-1)(x+1)= 0 \end{cases}$
`<=> x= -1`
`c)` Ta có: `2(x+1)^2≥0 => 2(x+1)^2+1≥1`
`=> 1/(2(x+1)^2+1)≤1`
`=> -1/(2(x+1)^2+1≥-1`
`C_min` `= -1 <=> 2(x+1)^2= 0 <=> x+1= 0 <=> x= -1`
