$a$) $(2x-1)^2 = |1-2x|$ ($1$)
$TH1$.$1-2x ≥ 0 ⇔ x ≤ \dfrac{1}{2}$
Từ ($1$) $⇒$ $(2x-1)^2 = 1-2x$
$⇔$ (2x-1)^2 - 1 + 2x = 0$
$⇔ 4x^2 - 4x + 1 - 1 + 2x = 0$
$⇔ 4x^2 - 2x = 0$
$⇔ 2x(2x-1) = 0$
$⇒$ \(\left[ \begin{array}{l}x=0(TM)\\x=\dfrac{1}{2}(TM)\end{array} \right.\)
$TH2$.$1-2x < 0 ⇔ x > \dfrac{1}{2}$
Từ ($1$) $⇒$ $(2x-1)^2 = 2x-1$
$⇔$ (2x-1)^2 - 2x + 1 = 0$
$⇔ 4x^2 - 4x + 1 -2x + 1= 0$
$⇔ 4x^2 - 6x + 2 = 0$
$⇔ 2x^2 - 3x + 1 = 0$
$⇔ 2x^2 - 2x - x + 1 = 0$
$⇔ 2x(x-1) - (x-1) = 0$
$⇔ (2x-1)(x-1)=0$
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{1}{2}(KTM)\\x=1(TM)\end{array} \right.\)
Vậy $S$ $=$ `{0;1/2;1}`
$b$) $(x-2)^2+(2x+1)^2=0$
Vì : $(x-2)^2;(2x+1)^2≥0$ $∀$ $x$
$⇒$ $\left \{ {{x=2} \atop {x=\dfrac{1}{2}}} \right.$ ($KTM$)
Vậy $S=∅$
