Bài 40 trang 23 SGK Toán 7 tập 1

Đề bài

Tính

a) ${\left( {\dfrac{3}{7} + \dfrac{1}{2}} \right)^2}$ 

b)  ${\left( {\dfrac{3}{4} - \dfrac{5}{6}} \right)^2}$

c) $\dfrac{{{5^4}{{.20}^4}}}{{{{25}^5}{{.4}^5}}}$

d) ${\left( {\dfrac{{ - 10}}{3}} \right)^5}.{\left( {\dfrac{{ - 6}}{5}} \right)^4}$

Lời giải chi tiết

a) ${\left( {\dfrac{3}{7} + \dfrac{1}{2}} \right)^2} = {\left( {\dfrac{6}{{14}} + \dfrac{7}{{14}}} \right)^2}$$\,= {\left( {\dfrac{{13}}{{14}}} \right)^2}= \dfrac{{169}}{{196}}$     

b)  ${\left( {\dfrac{3}{4} - \dfrac{5}{6}} \right)^2} = {\left( {\dfrac{9}{{12}} - \dfrac{{10}}{{12}}} \right)^2}$$\, = {\left( {\dfrac{{ - 1}}{{12}}} \right)^2} = \dfrac{1}{{144}}$

c)  $\dfrac{{{5^4}{{.20}^4}}}{{{{25}^5}{{.4}^5}}} = \dfrac{{{{\left( {5.20} \right)}^4}}}{{{{\left( {25.4} \right)}^5}}} = \dfrac{{{{100}^4}}}{{{{100}^5}}} = \dfrac{1}{{100}}$

d)  

$\begin{array}{l} {\left( {\dfrac{{ - 10}}{3}} \right)^5}.{\left( {\dfrac{{ - 6}}{5}} \right)^4}\\ = \left( {\dfrac{{ - 10}}{3}} \right).{\left( {\dfrac{{ - 10}}{3}} \right)^4}.{\left( {\dfrac{{ - 6}}{5}} \right)^4}\\ = \dfrac{{ - 10}}{3}.{\left[ {\left( {\dfrac{{ - 10}}{3}} \right).\left( {\dfrac{{ - 6}}{5}} \right)} \right]^4}\\ = \dfrac{{ - 10}}{3}.{\left[ {\dfrac{{ - 10.\left( { - 6} \right)}}{{3.5}}} \right]^4}\\ = \dfrac{{ - 10}}{3}.{\left( {\dfrac{{60}}{{15}}} \right)^4}\\ = \dfrac{{ - 10}}{3}{.4^4}\\ = \dfrac{{ - 10}}{3}.256 = \dfrac{{ - 2560}}{3} \end{array}$

Cách khác câu d:  

${\left( {\dfrac{{ - 10}}{3}} \right)^5}.{\left( {\dfrac{{ - 6}}{5}} \right)^4}$$= \dfrac{{{{\left( { - 10} \right)}^5}}}{{{3^5}}}.\dfrac{{{{\left( { - 6} \right)}^4}}}{{{5^4}}}$$\,= \dfrac{{{{\left( { - 2.5} \right)}^5}}}{{{3^5}}}.\dfrac{{{{\left( { - 2.3} \right)}^4}}}{{{5^4}}}$$= \dfrac{{{{\left( { - 2} \right)}^5}{{.5}^5}}}{{{3^5}}}.\dfrac{{{{\left( { - 2} \right)}^4}{{.3}^4}}}{{{5^4}}}$$= \dfrac{{  {(-2)^5}{{.5}^5}.{{(-2)}^4}{{.3}^4}}}{{{3^5}{{.5}^4}}} = \dfrac{{ {{5.(-2)}^9}}}{3} =  - \dfrac{{2560}}{3}$ 

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