\(\text{ }\dfrac{\left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-a\right)}{\left(b-a\right)\left(b-c\right)}.\left(x-c\right)+\dfrac{\left(x-b\right)}{\left(a-b\right)\left(a-c\right)}.\left(x-c\right)=1\)
\(\Leftrightarrow\left(x-c\right)\left(\dfrac{\left(x-a\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)}{\left(a-b\right)\left(a-c\right)}\right)=1\)
\(\Leftrightarrow\left(x-c\right)\dfrac{\left(a-x\right)\left(a-c\right)+\left(x-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\left(x-c\right)\left[\left(a^2-b^2\right)-x\left(a-b\right)-c\left(a-b\right)\right]=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow\left(x-c\right)\left(a-b\right)\left(a+b-x-c\right)=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow\left(x-c\right)\left(a+b-x-c\right)-\left(b-c\right)\left(a-c\right)=0\)
\(\Leftrightarrow ax+bx-x^2-xc-ac-bc+xc+c^2-ab+bc+ac-c^2=0\)
\(\Leftrightarrow x^2-ax-bx+ab=0\)
\(\Leftrightarrow x\left(x-a\right)+b\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(x-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=0\\x-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=a\\x=b\end{matrix}\right.\)
Vậy\(S=\left\{a,b\right\}\)
