a) \(\sqrt{9-12x+4x^2}=4\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.3+9}=4\Leftrightarrow\sqrt{\left(2x-3\right)^2}=4\left(1\right)\)Nếu \(x< \dfrac{3}{2}\)
\(\left(1\right)\Leftrightarrow3-2x=4\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)(nhận)
Nếu \(x\ge\dfrac{3}{2}\)
\(\left(1\right)\Leftrightarrow2x-3=4\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)(nhận)
Vậy S=\(\left\{\dfrac{-1}{2};\dfrac{7}{2}\right\}\)
b) \(\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}=1\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+1\right)^2}=1\left(1\right)\)Nếu x<-1
\(\left(1\right)\Leftrightarrow1-x+\left[-\left(x+1\right)\right]=1\Leftrightarrow1-x+\left(-x-1\right)=1\Leftrightarrow1-x-x-1=1\Leftrightarrow-2x=1\Leftrightarrow x=\dfrac{-1}{2}\)(loại)
Nếu -1≤x<1
\(\left(1\right)\Leftrightarrow1-x+x+1=1\Leftrightarrow2=1\)(loại)
Nếu x≥1
\(\left(1\right)\Leftrightarrow x-1+x+1=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)(loại)
Vậy S=∅
