Đáp án:

$\begin{array}{l}
c)\Delta ' = {\left( {a + b} \right)^2} - 4ab\\
 = {\left( {a - b} \right)^2}\\
 + Khi:\Delta ' = 0\\
 \Leftrightarrow a = b\\
 \Leftrightarrow x = \dfrac{a}{2} = \dfrac{b}{2}\\
 + Khi:a\# b\\
 \Leftrightarrow \left[ \begin{array}{l}
{x_1} = \dfrac{{ - b' + \sqrt {\Delta '} }}{a} = \dfrac{{a + b + \sqrt {{{\left( {a - b} \right)}^2}} }}{4}\\
{x_2} = \dfrac{{ - b' - \sqrt {\Delta '} }}{a} = \dfrac{{a + b - \sqrt {{{\left( {a - b} \right)}^2}} }}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
{x_1} = \dfrac{{a + b + \left| {a - b} \right|}}{4}\\
{x_2} = \dfrac{{a + b - \left| {a - b} \right|}}{4}
\end{array} \right.\\
 + Khi:a > b\\
 \Leftrightarrow \left[ \begin{array}{l}
{x_1} = \dfrac{{a + b + a - b}}{4} = \dfrac{a}{2}\\
{x_2} = \dfrac{{a + b - a + b}}{4} = \dfrac{b}{2}
\end{array} \right.\\
 + Khi:a < b\\
 \Leftrightarrow \left[ \begin{array}{l}
{x_1} = \dfrac{{a + b - a + b}}{4} = \dfrac{b}{2}\\
{x_2} = \dfrac{{a + b + a - b}}{4} = \dfrac{a}{2}
\end{array} \right.\\
Vậy\,x \in \left\{ {\dfrac{a}{2};\dfrac{b}{2}} \right\}
\end{array}$