1.3.27. Найдите значение выражения (4x)^2 * x^5 / (x^4 * 5x^3).

Привет! Давай решим эти задания по алгебре. **1.3.27.** $\frac{(4x)^2 \cdot x^5}{x^4 \cdot 5x^3} = \frac{16x^2 \cdot x^5}{5x^7} = \frac{16x^7}{5x^7} = \frac{16}{5} = 3.2$ **1.3.28.** $3^{x-4} = 9$ $3^{x-4} = 3^2$ $x - 4 = 2 \Rightarrow x = 6$ **1.3.29.** $4^{2-x} = 16$ $4^{2-x} = 4^2$ $2 - x = 2 \Rightarrow x = 0$ **1.3.30.** $4^{2x-17} = \frac{1}{64}$ $4^{2x-17} = 4^{-3}$ $2x - 17 = -3 \Rightarrow 2x = 14 \Rightarrow x = 7$ **1.3.31.** $36^{x-5} = \frac{1}{6}$ $(6^2)^{x-5} = 6^{-1}$ $2(x - 5) = -1 \Rightarrow 2x - 10 = -1 \Rightarrow 2x = 9 \Rightarrow x = 4.5$ **1.3.32.** $(\frac{1}{2})^{3x-12} = \frac{1}{8}$ $(2^{-1})^{3x-12} = 2^{-3}$ $-(3x - 12) = -3 \Rightarrow -3x + 12 = -3 \Rightarrow -3x = -15 \Rightarrow x = 5$ **1.3.33.** $(\frac{1}{9})^{x-7} = 3$ $(3^{-2})^{x-7} = 3^1$ $-2(x - 7) = 1 \Rightarrow -2x + 14 = 1 \Rightarrow -2x = -13 \Rightarrow x = 6.5$ **1.3.34.** $(\frac{1}{10})^{x-7} = 10^x$ $10^{-(x-7)} = 10^x$ $-x + 7 = x \Rightarrow 2x = 7 \Rightarrow x = 3.5$ **1.3.35.** $5^{7+2x} = 25^{2x}$ $5^{7+2x} = (5^2)^{2x}$ $7 + 2x = 4x \Rightarrow 2x = 7 \Rightarrow x = 3.5$ **1.3.36.** $3^{3+4x} = 1.5 \cdot 2^{3+4x}$ Разделим обе части на $2^{3+4x}$: $(\frac{3}{2})^{3+4x} = 1.5$ $1.5^{3+4x} = 1.5^1$ $3 + 4x = 1 \Rightarrow 4x = -2 \Rightarrow x = -0.5$ **1.3.37.** $(\frac{1}{2})^{x-6} = 32^x$ $(2^{-1})^{x-6} = (2^5)^x$ $-x + 6 = 5x \Rightarrow 6x = 6 \Rightarrow x = 1$ **1.3.38.** $(\frac{1}{2})^{x-5} = 2^x$ $(2^{-1})^{x-5} = 2^x$ $-x + 5 = x \Rightarrow 2x = 5 \Rightarrow x = 2.5$