loa loa loa..... có ai đó ko ạ , làm ơn giúp mình ít bài với...! Câu 1:tính: a, [tex]\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}[/tex] b,[tex]\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}[/tex] Câu 2:rút gọn: [tex]\frac{x-1}{x\sqrt{x}-1}[/tex] làm ơn giúp mình ,mình cm trước..
1a/ $\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}$ $= \sqrt{5 + 2\sqrt{5}.\sqrt{3} + 3} - \sqrt{5 - 2\sqrt{5}.\sqrt{3} + 3}$ $= \sqrt{(\sqrt5 + \sqrt3)^2} - \sqrt{(\sqrt5 - \sqrt3)^2}$ $= |\sqrt5 + \sqrt3| - |\sqrt5 - \sqrt3|$ $= \sqrt5 + \sqrt3 - \sqrt5 + \sqrt3$ $= 2\sqrt3$ b/ $\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}$ $= \dfrac{\sqrt{8 - 2\sqrt{15}}}{\sqrt{2}} - \dfrac{\sqrt{8+2\sqrt{5}}}{\sqrt{2}}$ $= -\dfrac{\sqrt{8+2\sqrt{5}} - \sqrt{8 - 2\sqrt{15}}}{\sqrt{2}}$ $= -\dfrac{2\sqrt{3}}{\sqrt{2}}$ $= -\sqrt{2}.\sqrt{3} = -\sqrt{6}$
2/ $\dfrac{x-1}{x\sqrt{x}-1}$ $= \dfrac{x-1}{(\sqrt{x})^3 - 1}$ $= \dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(x + \sqrt{x}+1)}$ $= \dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}$
