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

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) \({\left( {\dfrac{{ - 10}}{3}} \right)^5}.{\left( {\dfrac{{ - 6}}{5}} \right)^4} \)\(\,= \dfrac{{{{\left( { - 2.5} \right)}^5}}}{{{3^5}}}.\dfrac{{{{\left( { - 2.3} \right)}^4}}}{{{5^4}}} \)\(= \dfrac{{ - {2^5}{{.5}^5}{{.2}^4}{{.3}^4}}}{{{3^5}{{.5}^4}}} = \dfrac{{ - {{5.2}^9}}}{3} =  - \dfrac{{2560}}{3}\)


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