47. a^7 : a^3

Решение заданий: 47. $a^3 \cdot a^3 = a^{3+3} = a^6$ 48. $(2xy^2)^3 = 2^3 \cdot x^3 \cdot (y^2)^3 = 8x^3y^6$ 49. $a^7 : a^5 = a^{7-5} = a^2$ 50. $(\frac{1}{2}a^2)^{-3} = (\frac{1}{2})^{-3} \cdot (a^2)^{-3} = 2^3 \cdot a^{-6} = \frac{8}{a^6}$ 51. $\frac{a^{-5} \cdot a^2}{a^{-4}} = a^{-5+2 - (-4)} = a^{-5+2+4} = a^1 = a$ 52. $(\frac{x}{2})^{10} \cdot (\frac{x^2}{4})^{-4} = (\frac{x}{2})^{10} \cdot (\frac{x^2}{2^2})^{-4} = \frac{x^{10}}{2^{10}} \cdot (\frac{x}{2})^{-8} = \frac{x^{10}}{2^{10}} \cdot \frac{2^8}{x^8} = \frac{x^{10-8}}{2^{10-8}} = \frac{x^2}{2^2} = \frac{x^2}{4}$ 53. $\frac{x^2 + 4x + 4}{x+2} = \frac{(x+2)^2}{x+2} = x+2$ 54. $\frac{9a^2 - 6a + 1}{3a-1} = \frac{(3a-1)^2}{3a-1} = 3a-1$ 55. $\frac{24a^5b^7c^3}{36a^7b^4c^2} = \frac{24}{36} \cdot a^{5-7} \cdot b^{7-4} \cdot c^{3-2} = \frac{2}{3} a^{-2} b^3 c = \frac{2b^3c}{3a^2}$ 56. $\frac{4pq(p-q)}{2p^2(p^2-q^2)} = \frac{4pq(p-q)}{2p^2(p-q)(p+q)} = \frac{2q}{p(p+q)} = \frac{2q}{p^2+pq}$ 57. $\frac{a^2 - 64}{a+8} = \frac{(a-8)(a+8)}{a+8} = a-8$ 58. $\frac{121 - x^2}{x-11} = \frac{-(x^2 - 121)}{x-11} = \frac{-(x-11)(x+11)}{x-11} = -(x+11) = -x-11$ 59. $\frac{3+x}{4} + \frac{3x-7}{4} = \frac{3+x+3x-7}{4} = \frac{4x-4}{4} = \frac{4(x-1)}{4} = x-1$ 60. $\frac{a+4}{b} + \frac{2a}{b} = \frac{a+4+2a}{b} = \frac{3a+4}{b}$ 61. $\frac{2a+b}{a-2b} + \frac{a+3b}{2b-a} = \frac{2a+b}{a-2b} - \frac{a+3b}{a-2b} = \frac{2a+b-a-3b}{a-2b} = \frac{a-2b}{a-2b} = 1$ (при $a \neq 2b$) 62. Допущение: пример 62 выглядит как $8t + \frac{2}{t-3}$. Решение: $8t + \frac{2}{t-3} = \frac{8t(t-3)+2}{t-3} = \frac{8t^2-24t+2}{t-3}$