Giải bài 37 trang 22 - SGK Toán lớp 7 Tập 1

a) $\dfrac{4^2.4^3}{2^{10}} = \dfrac{4^5}{2^{10}} =\dfrac{(2^2)^5}{2^{10}} = \dfrac{2^{10}}{2^{10}} =1$

b) $\dfrac{(0,6)^5}{(0,2)^6} = \dfrac{(3.0,2)^5}{(0,2).(0,2)^5} = \dfrac{3^{5}.(0,2)^5}{(0,2).(0,2)^5} = \dfrac{3^{5}}{(0,2)} = \dfrac{243}{(0,2)} = 1215$

c) $\dfrac{2^7.9^3}{6^5.8^2} = \dfrac{2^2.2^5.(3^2)^3}{(2.3)^5.(2^3)^2} = \dfrac{2^2.2^5.3^6}{2^5.3^5.2^6} = \dfrac{3}{2^4} = \dfrac{3}{16}$

d) $\dfrac{6^3 + 3.6^2 + 3^3}{-13}$

$= \dfrac{(2.3)^3 + 3.(2.3)^2 + 3^3}{-13}$

$= \dfrac{2^3.3^3 + 2^2.3^3 + 3^3}{-13}$

$= \dfrac{3^3(2^3 + 2^2 + 1)}{-13}$

$= \dfrac{27.13}{-13} = -27$

Lưu ý:

$+ \,\,\,\, (x.y)^n = x^n.y^n\\ + \,\,\,\, (x : y)^n = x^n : y^n , \,\, (y \ne 0)$