104. Освободитесь от иррациональности в знаменателе дроби: 1) \frac{8}{\sqrt{13}} ; 2) \frac{48}{\sqrt{6}} ; 3) \frac{27}{2\sqrt{3}} ; 4) \frac{b⁵}{c√b} ; 5) \frac{a-7}{\sqrt{a}-7} ; 6) \frac{1}{\sqrt{33}+1} ; 7) \frac{41}{\sqrt{47}-\sqrt{6}} ; 8) \frac{17}{\sqrt{39}+\sqrt{5}} ; 9) \frac{x-7}{\sqrt{x+18}-5} ; 10) \frac{x²-9x}{\sqrt{x+7}-4} ; 11) \frac{x²-64}{4+\sqrt{x+8}} ; 12) \frac{m}{\sqrt{11+5m}-\sqrt{3m+11}} .

104. Освободитесь от иррациональности в знаменателе дроби:

1. $$\frac{8}{\sqrt{13}} = \frac{8 \cdot \sqrt{13}}{\sqrt{13} \cdot \sqrt{13}} = \frac{8\sqrt{13}}{13}$$

   Ответ: $$\frac{8\sqrt{13}}{13}$$

2. $$\frac{48}{\sqrt{6}} = \frac{48 \cdot \sqrt{6}}{\sqrt{6} \cdot \sqrt{6}} = \frac{48\sqrt{6}}{6} = 8\sqrt{6}$$

   Ответ: $$8\sqrt{6}$$

3. $$\frac{27}{2\sqrt{3}} = \frac{27 \cdot \sqrt{3}}{2\sqrt{3} \cdot \sqrt{3}} = \frac{27\sqrt{3}}{2 \cdot 3} = \frac{9\sqrt{3}}{2}$$

   Ответ: $$\frac{9\sqrt{3}}{2}$$

4. $$\frac{b^5}{c\sqrt{b}} = \frac{b^5 \cdot \sqrt{b}}{c\sqrt{b} \cdot \sqrt{b}} = \frac{b^5\sqrt{b}}{cb} = \frac{b^4\sqrt{b}}{c}$$

   Ответ: $$\frac{b^4\sqrt{b}}{c}$$

5. $$\frac{a-7}{\sqrt{a}-7} = \frac{(a-7)(\sqrt{a}+7)}{(\sqrt{a}-7)(\sqrt{a}+7)} = \frac{(a-7)(\sqrt{a}+7)}{a - 49}$$

   Ответ: $$\frac{(a-7)(\sqrt{a}+7)}{a - 49}$$

6. $$\frac{1}{\sqrt{33}+1} = \frac{1 \cdot (\sqrt{33}-1)}{(\sqrt{33}+1) (\sqrt{33}-1)} = \frac{\sqrt{33}-1}{33 - 1} = \frac{\sqrt{33}-1}{32}$$

   Ответ: $$\frac{\sqrt{33}-1}{32}$$

7. $$\frac{41}{\sqrt{47}-\sqrt{6}} = \frac{41(\sqrt{47}+\sqrt{6})}{(\sqrt{47}-\sqrt{6})(\sqrt{47}+\sqrt{6})} = \frac{41(\sqrt{47}+\sqrt{6})}{47 - 6} = \frac{41(\sqrt{47}+\sqrt{6})}{41} = \sqrt{47}+\sqrt{6}$$

   Ответ: $$\sqrt{47}+\sqrt{6}$$

8. $$\frac{17}{\sqrt{39}+\sqrt{5}} = \frac{17(\sqrt{39}-\sqrt{5})}{(\sqrt{39}+\sqrt{5})(\sqrt{39}-\sqrt{5})} = \frac{17(\sqrt{39}-\sqrt{5})}{39 - 5} = \frac{17(\sqrt{39}-\sqrt{5})}{34} = \frac{\sqrt{39}-\sqrt{5}}{2}$$

   Ответ: $$\frac{\sqrt{39}-\sqrt{5}}{2}$$

9. $$\frac{x-7}{\sqrt{x+18}-5} = \frac{(x-7)(\sqrt{x+18}+5)}{(\sqrt{x+18}-5)(\sqrt{x+18}+5)} = \frac{(x-7)(\sqrt{x+18}+5)}{x+18 - 25} = \frac{(x-7)(\sqrt{x+18}+5)}{x-7} = \sqrt{x+18}+5$$

   Ответ: $$\sqrt{x+18}+5$$

10. $$\frac{x^2-9x}{\sqrt{x+7}-4} = \frac{(x^2-9x)(\sqrt{x+7}+4)}{(\sqrt{x+7}-4)(\sqrt{x+7}+4)} = \frac{x(x-9)(\sqrt{x+7}+4)}{x+7 - 16} = \frac{x(x-9)(\sqrt{x+7}+4)}{x-9} = x(\sqrt{x+7}+4)$$

    Ответ: $$x(\sqrt{x+7}+4)$$

11. $$\frac{x^2-64}{4+\sqrt{x+8}} = \frac{(x^2-64)(4-\sqrt{x+8})}{(4+\sqrt{x+8})(4-\sqrt{x+8})} = \frac{(x-8)(x+8)(4-\sqrt{x+8})}{16-(x+8)} = \frac{(x-8)(x+8)(4-\sqrt{x+8})}{16-x-8} = \frac{(x-8)(x+8)(4-\sqrt{x+8})}{8-x} = \frac{-(8-x)(x+8)(4-\sqrt{x+8})}{8-x} = -(x+8)(4-\sqrt{x+8}) = (x+8)(\sqrt{x+8}-4)$$

    Ответ: $$(x+8)(\sqrt{x+8}-4)$$

12. $$\frac{m}{\sqrt{11+5m}-\sqrt{3m+11}} = \frac{m(\sqrt{11+5m}+\sqrt{3m+11})}{(\sqrt{11+5m}-\sqrt{3m+11})(\sqrt{11+5m}+\sqrt{3m+11})} = \frac{m(\sqrt{11+5m}+\sqrt{3m+11})}{11+5m - (3m+11)} = \frac{m(\sqrt{11+5m}+\sqrt{3m+11})}{11+5m - 3m - 11} = \frac{m(\sqrt{11+5m}+\sqrt{3m+11})}{2m} = \frac{\sqrt{11+5m}+\sqrt{3m+11}}{2}$$

    Ответ: $$\frac{\sqrt{11+5m}+\sqrt{3m+11}}{2}$$