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1) $\sqrt{(\sqrt{5}-3)^2}+\sqrt{(\sqrt{5}-2)^2}$ - Ta có: $\sqrt{(\sqrt{5}-3)^2} = |\sqrt{5}-3| = 3-\sqrt{5}$ (vì $\sqrt{5} < 3)$ - Ta cũng có: $\sqrt{(\sqrt{5}-2)^2} = |\sqrt{5}-2| = \sqrt{5}-2$ (vì $\sqrt{5} > 2)$ - Vậy: \[ \sqrt{(\sqrt{5}-3)^2}+\sqrt{(\sqrt{5}-2)^2} = (3-\sqrt{5}) + (\sqrt{5}-2) = 1 \] 2) $\sqrt{(3-\sqrt{10})^2}-\sqrt{(\sqrt{10}-5)^2}$ - Ta có: $\sqrt{(3-\sqrt{10})^2} = |3-\sqrt{10}| = \sqrt{10}-3$ (vì $3 < \sqrt{10})$ - Ta cũng có: $\sqrt{(\sqrt{10}-5)^2} = |\sqrt{10}-5| = 5-\sqrt{10}$ (vì $\sqrt{10} < 5)$ - Vậy: \[ \sqrt{(3-\sqrt{10})^2}-\sqrt{(\sqrt{10}-5)^2} = (\sqrt{10}-3) - (5-\sqrt{10}) = 2\sqrt{10}-8 \] 3) $\sqrt{(2-\sqrt{5})^2}-\sqrt{(2+\sqrt{5})^2}$ - Ta có: $\sqrt{(2-\sqrt{5})^2} = |2-\sqrt{5}| = \sqrt{5}-2$ (vì $2 < \sqrt{5})$ - Ta cũng có: $\sqrt{(2+\sqrt{5})^2} = |2+\sqrt{5}| = 2+\sqrt{5}$ (vì $2+\sqrt{5} > 0)$ - Vậy: \[ \sqrt{(2-\sqrt{5})^2}-\sqrt{(2+\sqrt{5})^2} = (\sqrt{5}-2) - (2+\sqrt{5}) = -4 \] 4) $\sqrt{(3+\sqrt{5})^2}+\sqrt{(\sqrt{5}-6)^2}$ - Ta có: $\sqrt{(3+\sqrt{5})^2} = |3+\sqrt{5}| = 3+\sqrt{5}$ (vì $3+\sqrt{5} > 0)$ - Ta cũng có: $\sqrt{(\sqrt{5}-6)^2} = |\sqrt{5}-6| = 6-\sqrt{5}$ (vì $\sqrt{5} < 6)$ - Vậy: \[ \sqrt{(3+\sqrt{5})^2}+\sqrt{(\sqrt{5}-6)^2} = (3+\sqrt{5}) + (6-\sqrt{5}) = 9 \] 5) $\sqrt{(\sqrt{5}-1)^2}.(\sqrt{5}+1)$ - Ta có: $\sqrt{(\sqrt{5}-1)^2} = |\sqrt{5}-1| = \sqrt{5}-1$ (vì $\sqrt{5} > 1)$ - Vậy: \[ \sqrt{(\sqrt{5}-1)^2}.(\sqrt{5}+1) = (\sqrt{5}-1).(\sqrt{5}+1) = (\sqrt{5})^2 - 1^2 = 5 - 1 = 4 \]