Question

Đã trả lời bởi chuyên gia

Answer 1

$1, a, A = \sqrt{19 + 8 \sqrt{3}} - \sqrt{28 - \sqrt{300}} - \sqrt{12} = \sqrt{16 + 2.4 . \sqrt{3} + 3} - \sqrt{28 - 10 \sqrt{3}} - \sqrt{4.3} = \sqrt{{(4 + \sqrt{3})}^{2}} - \sqrt{25 - 2.5 . \sqrt{3} + 3} - 2 \sqrt{3} = |4 + \sqrt{3}| - \sqrt{{(5 - \sqrt{3})}^{2}} - 2 \sqrt{3} = 4 + \sqrt{3} - |5 - \sqrt{3}| - 2 \sqrt{3} (d o 4 + \sqrt{3} > 0) = 4 - \sqrt{3} - 5 + \sqrt{3} (d o 5 - \sqrt{3} > 0) = - 1$

Vậy A=-1

$b, B = \frac{(\sqrt{10} + \sqrt{6}) . \sqrt{4 - \sqrt{15}}}{8 - 2 \sqrt{15}} - \sqrt{15} = \frac{(\sqrt{10} + \sqrt{6}) . \sqrt{4 - \sqrt{5}}}{2 . (4 - \sqrt{15})} - \sqrt{15} = \frac{\sqrt{10} + \sqrt{6}}{2 \sqrt{4 - \sqrt{15}}} - \sqrt{15} = \frac{\sqrt{10} + \sqrt{6}}{\sqrt{2} . \sqrt{8 - 2 \sqrt{15}}} - \sqrt{15} = \frac{\sqrt{2} . (\sqrt{5} + \sqrt{3})}{\sqrt{2} . \sqrt{5 - 2 . \sqrt{5} . \sqrt{3} + 3}} - \sqrt{15} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{{(\sqrt{5} - \sqrt{3})}^{2}}} - \sqrt{15} = \frac{\sqrt{5} + \sqrt{3}}{|\sqrt{5} - \sqrt{3}|} - \sqrt{15} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \sqrt{15} (d o \sqrt{5} - \sqrt{3} > 0) = \frac{{(\sqrt{5} + \sqrt{3})}^{2}}{(\sqrt{5} - \sqrt{3}) (\sqrt{5} + \sqrt{3})} - \sqrt{15} = \frac{(8 + 2 \sqrt{15})}{5 - 3} - \sqrt{15} = 4 + \sqrt{15} - \sqrt{15} = 4$

Vậy B=4

Bài 2:

$a, \sqrt{2 x - 1} = 5, đ k : x \geq \frac{1}{2} < = > 2 x - 1 = 25 < = > 2 x = 26 < = > x = 13 (t m) V ậ t x = 13 b, \sqrt{x^{2} - 4 x + 2} - \sqrt{x - 2} = 0, đ k : x \geq 2 < = > \sqrt{x^{2} - 4 x + 2} = \sqrt{x - 2} < = > x^{2} - 4 x + 2 = x - 2 < = > x^{2} - 5 x + 4 = 0 < = > (x^{2} - 4 x) - (x - 4) = 0 < = > x . (x - 4) - (x - 4) = 0 < = > (x - 4) (x - 1) = 0 < = > x - 4 = 0 h o ặ c x - 1 = 0 < = > x = 4 (t . m) h o ặ c x = 1 (l o ạ i) V ậ y x = 4$

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Answer 2

$1, a, A = \sqrt{19 + 8 \sqrt{3}} - \sqrt{28 - \sqrt{300}} - \sqrt{12} = \sqrt{16 + 2.4 . \sqrt{3} + 3} - \sqrt{28 - 10 \sqrt{3}} - \sqrt{4.3} = \sqrt{{(4 + \sqrt{3})}^{2}} - \sqrt{25 - 2.5 . \sqrt{3} + 3} - 2 \sqrt{3} = |4 + \sqrt{3}| - \sqrt{{(5 - \sqrt{3})}^{2}} - 2 \sqrt{3} = 4 + \sqrt{3} - |5 - \sqrt{3}| - 2 \sqrt{3} (d o 4 + \sqrt{3} > 0) = 4 - \sqrt{3} - 5 + \sqrt{3} (d o 5 - \sqrt{3} > 0) = - 1$

Vậy A=-1

$b, B = \frac{(\sqrt{10} + \sqrt{6}) . \sqrt{4 - \sqrt{15}}}{8 - 2 \sqrt{15}} - \sqrt{15} = \frac{(\sqrt{10} + \sqrt{6}) . \sqrt{4 - \sqrt{5}}}{2 . (4 - \sqrt{15})} - \sqrt{15} = \frac{\sqrt{10} + \sqrt{6}}{2 \sqrt{4 - \sqrt{15}}} - \sqrt{15} = \frac{\sqrt{10} + \sqrt{6}}{\sqrt{2} . \sqrt{8 - 2 \sqrt{15}}} - \sqrt{15} = \frac{\sqrt{2} . (\sqrt{5} + \sqrt{3})}{\sqrt{2} . \sqrt{5 - 2 . \sqrt{5} . \sqrt{3} + 3}} - \sqrt{15} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{{(\sqrt{5} - \sqrt{3})}^{2}}} - \sqrt{15} = \frac{\sqrt{5} + \sqrt{3}}{|\sqrt{5} - \sqrt{3}|} - \sqrt{15} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \sqrt{15} (d o \sqrt{5} - \sqrt{3} > 0) = \frac{{(\sqrt{5} + \sqrt{3})}^{2}}{(\sqrt{5} - \sqrt{3}) (\sqrt{5} + \sqrt{3})} - \sqrt{15} = \frac{(8 + 2 \sqrt{15})}{5 - 3} - \sqrt{15} = 4 + \sqrt{15} - \sqrt{15} = 4$

Vậy B=4

Bài 2:

$a, \sqrt{2 x - 1} = 5, đ k : x \geq \frac{1}{2} < = > 2 x - 1 = 25 < = > 2 x = 26 < = > x = 13 (t m) V ậ t x = 13 b, \sqrt{x^{2} - 4 x + 2} - \sqrt{x - 2} = 0, đ k : x \geq 2 < = > \sqrt{x^{2} - 4 x + 2} = \sqrt{x - 2} < = > x^{2} - 4 x + 2 = x - 2 < = > x^{2} - 5 x + 4 = 0 < = > (x^{2} - 4 x) - (x - 4) = 0 < = > x . (x - 4) - (x - 4) = 0 < = > (x - 4) (x - 1) = 0 < = > x - 4 = 0 h o ặ c x - 1 = 0 < = > x = 4 (t . m) h o ặ c x = 1 (l o ạ i) V ậ y x = 4$

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