Bài 41 trang 23 SGK Toán 7 tập 1

Đề bài

Tính:

a) $\left( {1 + \dfrac{2}{3} - \dfrac{1}{4}} \right).{\left( {\dfrac{4}{5} - \dfrac{3}{4}} \right)^2}$

b) $2:{\left( {\dfrac{1}{2} - \dfrac{2}{3}} \right)^3}$

Lời giải chi tiết

a)

$\begin{array}{l} \left( {1 + \dfrac{2}{3} - \dfrac{1}{4}} \right).{\left( {\dfrac{4}{5} - \dfrac{3}{4}} \right)^2} \\= \left( {\dfrac{{12}}{{12}} + \dfrac{8}{{12}} - \dfrac{3}{{12}}} \right).{\left( {\dfrac{{16}}{{20}} - \dfrac{{15}}{{20}}} \right)^2}\\  = \dfrac{{17}}{{12}}.{\left( {\dfrac{1}{{20}}} \right)^2} =\dfrac{{17}}{{12}}.\dfrac{{{1^2}}}{{{{20}^2}}}\\ = \dfrac{{17}}{{12}}.\dfrac{1}{{400}} = \dfrac{{17}}{{4800}} \end{array}$

b)   $2:{\left( {\dfrac{1}{2} - \dfrac{2}{3}} \right)^3} = 2:{\left( {\dfrac{3}{6} - \dfrac{4}{6}} \right)^3}$

$= 2:{\left( { - \dfrac{1}{6}} \right)^3} = 2:\left( { - \dfrac{1}{{216}}} \right)$

$=  2.(-216) =  - 432$

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