Đáp án:
Giải thích các bước giải:
1) $ \sqrt[]{(x - 1)²} = |x - 1| = x - 1 ( x ≥ 1)$
2) $ \sqrt[]{(x - 2)²} = |x - 2| = - (x - 2) = 2 - x ( x < 2)$
3) $ \sqrt[]{a^{6}} = \sqrt[]{(a³)²} = |a³| = - a³ ( a < 0)$
4) $ \sqrt[]{a^{4}} = \sqrt[]{(a²)²} = |a²| = a² ( ∀a)$
5) $ \sqrt[]{64a²} + 2a = \sqrt[]{(8a)²} + 2a = |8a| + 2a = 8a + 2a = 10a (a ≥ 0)$
6) $ 3\sqrt[]{9a^{6}} - 6a³ = 3\sqrt[]{(3a³)²} - 6a³ = 3|3a³| - 6a³ = - 9a³ - 6a³ = - 15a³ ( a < 0)$
7) $ \sqrt[]{a² + 6a + 9} + \sqrt[]{a² - 6a + 9} = \sqrt[]{(a + 3)²} + \sqrt[]{(a - 3)²} = |a + 3| + |a - 3|$
8) $ \sqrt[]{a + 2\sqrt[]{a - 1}} + \sqrt[]{a - 2\sqrt[]{a - 1}} = \sqrt[]{a - 1 + 2\sqrt[]{a - 1} + 1} + \sqrt[]{a - 1 - 2\sqrt[]{a - 1} + 1} $
$ = \sqrt[]{(\sqrt[]{a - 1} + 1)²} + \sqrt[]{(\sqrt[]{a - 1} - 1)²} = |\sqrt[]{a - 1} + 1| + |\sqrt[]{a - 1} - 1|$
$ = (\sqrt[]{a - 1} + 1) - (\sqrt[]{a - 1} - 1) = 2 ( 1 ≤ a ≤ 2)$
9) $\frac{a\sqrt[]{a} - 8 + 2a - 4\sqrt[]{a}}{4 - a} = \frac{(a - 4)\sqrt[]{a} + 2(a - 4)}{4 - a} = \frac{(a - 4)(\sqrt[]{a} + 2)}{4 - a} = \sqrt[]{a} + 2 ( a ≥ 0; a \neq4 )$
10) $ 2x - \sqrt[]{4x² - 4x + 1} = 2x - \sqrt[]{(2x - 1)²} = 2x - |2x - 1| = 2x - (2x - 1) = 1 ( x > \frac{1}{2}$
11) $ \sqrt[]{x² + 4x + 4} + \sqrt[]{x²} = \sqrt[]{(x + 2)²} + \sqrt[]{x²} = |x + 2| + |x| = (x + 2) + x = 2(x + 1)( x ≥ 0)$
12) $ \sqrt[]{x + 2\sqrt[]{x - 1}} - \sqrt[]{x - 1} + 4 = \sqrt[]{x - 1 + 2\sqrt[]{x - 1} + 1} - \sqrt[]{x - 1} + 4$
$ = \sqrt[]{(\sqrt[]{x - 1} + 1)²} - \sqrt[]{x - 1} + 4 = |\sqrt[]{x - 1} + 1| - \sqrt[]{x - 1} + 4 $
$ = (\sqrt[]{x - 1} + 1) - \sqrt[]{x - 1} + 4 = 5 ( x ≥ 1)$
13) $ |x - 2| + \frac{\sqrt[]{x² - 4x + 4}}{x - 2} = |x - 2| + \frac{\sqrt[]{(x - 2)²}}{x - 2}$
$= |x - 2| + \frac{|x -2|}{x - 2} = (x - 2) + \frac{x - 2}{x - 2} = (x - 2) + 1 = x - 1 ( x > 2)$
14) $|4 - x| + \frac{4 - x}{\sqrt[]{x² - 8x + 16}} = |4 - x| + \frac{4 - x}{\sqrt[]{(4 - x)²}}$
$ = |4 - x| + \frac{4 - x}{|4 - x|} = (4 - x) + \frac{4 - x}{4 - x} = (4 - x) + 1 = 5 - x ( x < 4)$
