b) $\sqrt[]{x² - 2x - 4}$ = $\sqrt[]{x-2}$
ĐKXĐ: x ≥ 1+$\sqrt[]{5}$ hoặc x ≤ 1 - $\sqrt[]{5}$
⇔ x² - 2x - 4 = 2 - x
⇔ x² - x - 6 = 0
⇔ x² - 3x + 2x - 6 = 0
⇔ x(x-3) + 2(x-3) = 0
⇔ (x - 3)(x + 2) = 0
⇔ \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
d) $\sqrt[]{x² - 2x + 1}$ = x-1
ĐKXĐ: x ≥ 1
⇔ $\sqrt[]{(x - 1)²}$ = x - 1
⇔ |x - 1| = x - 1
⇔ x - 1 ≥ 0
⇔ x ≥ 1 thỏa mãn
f) $\sqrt[]{25x² - 10x + 1}$ = 5x - 1
ĐKXĐ: x ≥ $\frac{1}{5}$
⇔ $\sqrt[]{(5x - 1)²}$ = 5x - 1
⇔ |5x - 1| = 5x - 1
⇔ 5x - 1 ≥ 0
⇔ x ≥ $\frac{1}{5}$
