Giải bài 52 trang 26 SBT Toán 7 Cánh diều tập 1 - CD

Phương pháp giải:

Ta thực hiện phép tính bằng cách tính nhanh.

Lưu ý:

      \(\dfrac{{{x^m}}}{{{y^m}}} = {\left( {\dfrac{x}{y}} \right)^m}{\rm{ }}(y \ne 0)\).

     \({(x.y)^m} = {x^m}.{y^m}\).

     \({x^m}.{x^n} = {x^{m + n}};{\rm{ }}\dfrac{{{x^m}}}{{{x^n}}} = {x^{m - n}}{\rm{ (}}m > n)\).

Lời giải chi tiết:

a)

\(\dfrac{{ - 5}}{7}.\dfrac{2}{{11}} + \dfrac{{ - 5}}{7}.\dfrac{9}{{11}} + \dfrac{5}{7} = \dfrac{{ - 5}}{7}.\left( {\dfrac{2}{{11}} + \dfrac{9}{{11}}} \right) + \dfrac{5}{7} = \dfrac{{ - 5}}{7}.1 + \dfrac{5}{7} = \dfrac{{ - 5}}{7} + \dfrac{5}{7} = 0\)

b)

 \(\begin{array}{l}\left[ {\left( {\dfrac{{ - 3}}{8} + \dfrac{{11}}{{23}}} \right):\dfrac{5}{9} + \left( {\dfrac{{ - 5}}{8} + \dfrac{{12}}{{23}}} \right):\dfrac{5}{9}} \right].\dfrac{{ - 11}}{{325}} = \left[ {\left( {\dfrac{{ - 3}}{8} + \dfrac{{11}}{{23}} + \dfrac{{ - 5}}{8} + \dfrac{{12}}{{23}}} \right):\dfrac{5}{9}} \right].\dfrac{{ - 11}}{{325}}\\{\rm{   }} = \left[ {( - 1 + 1):\dfrac{5}{9}} \right].\dfrac{{ - 11}}{{325}} = \left( {0:\dfrac{5}{9}} \right).\dfrac{{ - 11}}{{325}} = 0.\dfrac{{ - 11}}{{325}} = 0\end{array}\)

c*)

\(\dfrac{{{{15}^5}}}{{{5^5}}} - {( - 0,25)^2}{.4^2} = {\left( {\dfrac{{15}}{5}} \right)^5} - {\left( {\dfrac{{ - 1}}{4}} \right)^2}{.4^2} = {3^5} - \dfrac{1}{{{4^2}}}{.4^2} = {3^5} - 1 = 243 - 1 = 242\)

d*)

 \(\begin{array}{l} - \dfrac{{{2^{15}}{{.9}^4}}}{{{6^6}{{.8}^3}}} + 0,75.\dfrac{{ - 1}}{2} + 0,375 =  - \dfrac{{{2^{15}}{{.3}^{{2^4}}}}}{{{{(2.3)}^6}{{.2}^{{3^3}}}}} + ( - 0,375) + 0,375\\{\rm{  }} =  - \dfrac{{{2^{15}}{{.3}^{{2^4}}}}}{{{{(2.3)}^6}{{.2}^{{3^3}}}}} + \left[ {( - 0,375) + 0,375} \right] =  - \dfrac{{{2^{15}}{{.3}^{{2^4}}}}}{{{{(2.3)}^6}{{.2}^{{3^3}}}}}\\{\rm{  }} =  - \dfrac{{{2^{15}}{{.3}^8}}}{{{2^6}{{.3}^6}{{.2}^9}}} =  - \dfrac{{{2^{15}}{{.3}^8}}}{{{2^{15}}{{.3}^6}}} =  - \dfrac{{{3^8}}}{{{3^6}}} =  - {3^2} =  - 9\end{array}\). 

-- Mod Toán 7 HỌC247