Đáp án:
Giải thích các bước giải:
1.
d) Để biểu thức đã cho có nghĩa thì:
(x - 1)(x - 3) ≥ 0
⇔ \(\left[ \begin{array}{l}\left \{ {{x-1≥0} \atop {x-3≥0}} \right.\\\left \{ {{x-1≤0} \atop {x-3≤0}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\left \{ {{x≥1} \atop {x≥3}} \right.\\\left \{ {{x≤1} \atop {x≤3}} \right.\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x≥3\\x≤1\end{array} \right.\)
e) Để biểu thức đã cho có nghĩa thì:
$\left \{ {{x+4≥0} \atop {x-1≥0}} \right.$
⇔ $\left \{ {{x≥-4} \atop {x≥1}} \right.$
⇔ x ≥ 1
2.
a) $3\sqrt{50} - 2\sqrt{12} - \sqrt{18} + \sqrt{75} - \sqrt{8}$
= $3×5\sqrt{2} - 2×2\sqrt{3} - 3\sqrt{2} + 5\sqrt{3} - 2\sqrt{2}$
= $10\sqrt{2} - 4\sqrt{3} - 3\sqrt{2} + 5\sqrt{3} - 2\sqrt{2}$
= $5\sqrt{2} + \sqrt{3}$
b) $2\sqrt{75} - 3\sqrt{12} + \sqrt{27}$
= $2×5\sqrt{3} - 3×2\sqrt{3} + 3\sqrt{3}$
= $10\sqrt{3} - 6\sqrt{3} + 3\sqrt{3}$
= $7\sqrt{3}$
c) $\sqrt{27} - \sqrt{12} + \sqrt{75} + \sqrt{147}$
= $3\sqrt{3} - 2\sqrt{3} + 5\sqrt{3} + 7\sqrt{3}$
= $13\sqrt{3}$
d) $\sqrt{17 - 4\sqrt{9 + 4\sqrt{5}}}$
= $\sqrt{17 - 4\sqrt{2² + 2×2×\sqrt{5} + \sqrt{5}²}}$
= $\sqrt{17 - 4\sqrt{(2 + \sqrt{5})²}}$
= $\sqrt{17 - 4|2 + \sqrt{5}|}$
= $\sqrt{17 - 8 - 4\sqrt{5}}$ (vì $2 + \sqrt{5} > 0$)
= $\sqrt{9 - 4\sqrt{5}}$
= $\sqrt{\sqrt{5}² - 2×2×\sqrt{5} + 2²}$
= $\sqrt{(\sqrt{5} - 2)²}$
= $|\sqrt{5} - 2|$
= $\sqrt{5} - 2$ (vì $\sqrt{5} - 2 > 0$)
e) $\sqrt{13 + 30\sqrt{2 + \sqrt{9 + 4\sqrt{2}}}}$
= $\sqrt{13 + 30\sqrt{2 + \sqrt{1² + 2×2×\sqrt{2} + (2\sqrt{2})²}}}$
= $\sqrt{13 + 30\sqrt{2 + \sqrt{(1 + 2\sqrt{2})²}}}$
= $\sqrt{13 + 30\sqrt{2 + |1 + 2\sqrt{2}|}}$
= $\sqrt{13 + 30\sqrt{2 + 1 + 2\sqrt{2}}}$ (vì 1 + 2$\sqrt{2} > 0$)
= $\sqrt{13 + 30\sqrt{3 + 2\sqrt{2}}}$
= $\sqrt{13 + 30\sqrt{1² + 2\sqrt{2} + \sqrt{2}²}}$
= $\sqrt{13 + 30\sqrt{(1 + \sqrt{2})²}}$
= $\sqrt{13 + 30|1 + \sqrt{2}|}$
= $\sqrt{13 + 30 + 30\sqrt{2}}$ (vì $1 + \sqrt{2} > 0$)
= $\sqrt{43 + 30\sqrt{2}}$
= $\sqrt{5² + 5×3\sqrt{2} + (3\sqrt{2})²}$
= $\sqrt{(5 + 3\sqrt{2})²}$
= $|5 + 3\sqrt{2}|$
= $5 + 3\sqrt{2}$ (vì 5 + 3$\sqrt{2} > 0$)
