Bài 5:
a) $\sqrt{6} + \sqrt{3}$ và $\sqrt{5} + \sqrt{6}$
Ta có:
$(\sqrt{7} + \sqrt{3})^2 = 7 + 2\sqrt{7.3} + 3 = 10 + 2\sqrt{21}$
$(\sqrt{5} + \sqrt{6})^2 = 5 + 2\sqrt{5.6} + 6 = 11 + 2\sqrt{30}$
Do $11 + 2\sqrt{30} > 10 + 2\sqrt{21}$
nên $(\sqrt{5} + \sqrt{6})^2 > (\sqrt{7} + \sqrt{3})^2$
hay $\sqrt{5} + \sqrt{6} > \sqrt{6} + \sqrt{3}$
b) $\sqrt{4 - 3\sqrt{3}}$ và $\sqrt{3} - 1$
Ta có: $\sqrt{3} - 1 = \sqrt{(\sqrt{3} - 1)^2} = \sqrt{4 - 2\sqrt{3}}$
Do $4 - 2\sqrt{3} > 4 - 3\sqrt{3}$
nên $\sqrt{4 - 2\sqrt{3}} > \sqrt{4 - 3\sqrt{3}}$
hay $\sqrt{3} - 1 > \sqrt{4 - 3\sqrt{3}}$
Bài 6:
$\begin{array}{l}a. \, A = x - y - 3(\sqrt{x} + \sqrt{y})\\=(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) - 3(\sqrt{x} + \sqrt{y})\\=(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y} - 3)\\b. \, B = x - 4\sqrt{x} + 4\\=(\sqrt{x})^2 - 2.2.\sqrt{x} + 2^2\\=(\sqrt{x} - 2)^2\\c. \, C = \sqrt{x^3} - \sqrt{y}^3 + \sqrt{x^2y} - \sqrt{xy^2}\\=(\sqrt{x} - \sqrt{y})(\sqrt{x^2} + \sqrt{xy} + \sqrt{y^2})+\sqrt{xy}(\sqrt{x} - \sqrt{y})\\=(\sqrt{x} - \sqrt{y})(\sqrt{x^2} + \sqrt{xy} + \sqrt{y^2} + \sqrt{xy})\\=(\sqrt{x} - \sqrt{y})(\sqrt{x^2} + 2\sqrt{xy} + \sqrt{y^2})\\=(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})^2\\d. \, D = 5x^2 - 7x\sqrt{y} + 2y\\=5x^2 - 5x\sqrt{y} - 2x\sqrt{y} + 2y\\=5x(x-\sqrt{y}) - 2\sqrt{y}(x - \sqrt{y})\\=(x - 2\sqrt{y})(5x - 2\sqrt{y})\end{array}$
