# Bài 37 trang 22 SGK Toán 7 tập 1

##### Hướng dẫn giải

Chú ý các công thức sau:

$$\begin{array}{l} {\left( {x.y} \right)^n} = {x^n}.{y^n}\\ {\left( {\frac{x}{y}} \right)^n} = \frac{{{x^n}}}{{{y^n}}}\,\,\left( {y \ne 0} \right) \end{array}$$

$${\left( {{x^n}} \right)^m} = {x^{n.m}}$$

Lời giải chi tiết

a)   $$\frac{4^{2}.4^{3}}{2^{10}}$$ = $$\frac{4^{5}}{(2^{2})^{5}}=\frac{4^{5}}{4^{5}}= 1$$

b) $$\frac{(0,6)^{5}}{(0,2)^{6}}$$ = $$\frac{(0,2.3)^{5}}{(0,2)^{6}} = \frac{(0,2)^{5}.3^{5}}{(0,2)^{5}.0,2} = \frac{3^{5}}{0,2} = \frac{243}{0,2}= 1215$$

c)

$$\frac{{{2^7}{{.9}^3}}}{{{6^5}{{.8}^2}}} = \frac{{{2^7}.{{\left( {{3^2}} \right)}^3}}}{{{{\left( {2.3} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} = \frac{{{2^7}{{.3}^6}}}{{{2^5}{{.3}^5}{{.2}^6}}} = \frac{{{2^7}{{.3}^6}}}{{{2^{11}}{{.3}^5}}}$$$$= \frac{3}{{{2^4}}} = \frac{3}{{16}}$$

(Áp dụng công thức: $${\left( {{x^n}} \right)^m} = {x^{n.m}};\,\,{\left( {x.y} \right)^n} = {x^n}.{y^n}$$)

d) $$\frac{6^{3} + 3.6^{2}+ 3^{3}}{-13}= {{{{\left( {2.3} \right)}^3} + 3.{{\left( {2.3} \right)}^2} + {3^3}} \over { - 13}}$$$$=\frac{2^{3}.3^{3} + 3^{3}.2^{2} + 3^{3}}{-13} = \frac{3^{3}.(2^{3} + 2^{2} + 1)}{-13}$$$$= -3^{3} = -27$$