Bài 37 trang 22 SGK Toán 7 tập 1

Đề bài

Tìm giá trị của biểu thức sau

a) $\dfrac{4^{2}.4^{3}}{2^{10}}$  

b) $\dfrac{(0,6)^{5}}{(0,2)^{6}}$

c)$\dfrac{2^{7}. 9^{3}}{6^{5}.8^{2}}$

d) $\dfrac{6^{3} + 3.6^{2}+ 3^{3}}{-13}$

Lời giải chi tiết

a)   $\dfrac{4^{2}.4^{3}}{2^{10}} = \dfrac{4^{2+3}}{2^{2.5}}$$= \dfrac{4^{5}}{(2^{2})^{5}}=\dfrac{4^{5}}{4^{5}}= 1$

b) $\dfrac{(0,6)^{5}}{(0,2)^{6}} = \dfrac{(0,2.3)^{5}}{(0,2)^{5+1}} = \dfrac{(0,2)^{5}.3^{5}}{(0,2)^{5}.0,2}$

$= \dfrac{3^{5}}{0,2} = \dfrac{243}{0,2}= 1215$

c) $\dfrac{{{2^7}{{.9}^3}}}{{{6^5}{{.8}^2}}} = \dfrac{{{2^7}.{{\left( {{3^2}} \right)}^3}}}{{{{\left( {2.3} \right)}^5}.{{\left( {{2^3}} \right)}^2}}}$$= \dfrac{{{2^7}{{.3}^{2.3}}}}{{{2^5}{{.3}^5}{{.2}^{3.2}}}} = \dfrac{{{2^7}{{.3}^6}}}{{{2^5}{{.3}^5}{{.2}^6}}}$

$= \dfrac{{{2^7}{{.3}^6}}}{{{2^{11}}{{.3}^5}}} = \dfrac{3}{{{2^4}}} = \dfrac{3}{{16}}$

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

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

Cách khác câu d:

$\begin{array}{l} \dfrac{{{6^3} + {{3.6}^2} + {3^3}}}{{ - 13}}\\ = \dfrac{{216 + 108 + 27}}{{ - 13}}\\ = \dfrac{{351}}{{ - 13}} = - 27 \end{array}$

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