Bài 40 trang 23 SGK Toán 7 tập 1

Đề bài

Tính

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

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

c) \(\frac{{{5^4}{{.20}^4}}}{{{{25}^5}{{.4}^5}}}\)

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

Hướng dẫn giải

Ta sử dụng các công thức sau:

\({\left( {x.y} \right)^n} = {x^n}.{y^n}\)

\({x^n} = \underbrace {x.x.x...x}_{n\,\,\,so}\left( {x \in Q,n \in N,n > 1} \right)\)

\({\left( {\frac{x}{y}} \right)^n} = \frac{{{x^n}}}{{{y^n}}}\,\,\left( {y \ne 0} \right)\)

Lời giải chi tiết

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

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

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

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