Bài 24: So sánh:a)$(-0.1)^{2}$$\times$ $(-0.1)^{4}$ và$[(-...

a) Ta có $(-0.1)^2 \times (-0.1)^4 = (-0.1)^{2+4} = (-0.1)^6 = (-0.1)^{3 \times 2} = [(-0.1)^3]^2$. Vậy $(-0.1)^2 \times (-0.1)^4 = [(-0.1)^3]^2$.

b) Ta có $(\frac{1}{2})^8 / (\frac{1}{2})^2 = (\frac{1}{2})^{8-2} = (\frac{1}{2})^6 = (\frac{1}{2})^{3+3} = (\frac{1}{2})^3 \times (\frac{1}{2})^3$. Vậy $(\frac{1}{2})^8 / (\frac{1}{2})^2 = (\frac{1}{2})^3 \times (\frac{1}{2})^3$.

c) Ta có $9^8 / 27^3 = (3^2)^8 / (3^3)^3 = 3^{2 \times 8} / 3^{3 \times 3} = 3^{16} / 3^9 = 3^7 = 3^{2+5} = 3^2 \times 3^5$. Vậy $9^8 / 27^3 = 3^2 \times 3^5$.

d) Ta có $(\frac{1}{4})^7 \times 0.25 = (\frac{1}{4})^7 \times \frac{1}{4} = (\frac{1}{4})^{7+1} = (\frac{1}{4})^8 = (\frac{1}{4})^{2 \times 4} = [(\frac{1}{4})^2]^4$. Vậy $(\frac{1}{4})^7 \times 0.25 = [(\frac{1}{4})^2]^4.

e) Ta có $[(-0.7)^2]^3 = [(0.7)^2]^3 = (0.7)^{2 \times 3} = (0.7)^{3 \times 2} = [(-0.7)^3]^2$. Vậy $[(-0.7)^2]^3 = [(-0.7)^3]^2$.