Một cái sân hình chữ nhật có diện tích \(\dfrac{{77}}{5}\,{m^2},\) chiều dài là \(\dfrac{{22}}{5}m\). Chu vi của cái sân đó là:
Chiều rộng của cái sân là: \(\dfrac{{77}}{5}:\dfrac{{22}}{5} = \dfrac{{77}}{5}.\dfrac{5}{{22}} = \dfrac{7}{2}\,\left( m \right)\)
Chu vi của cái sân là: \(2.\left( {\dfrac{{22}}{5} + \dfrac{7}{2}} \right) = \dfrac{{79}}{5}\,\left( m \right).\)
Cho \(13\dfrac{3}{x} = \dfrac{{136}}{{10}}\). Giá trị của \(x\) là:
\(\begin{array}{l}13\dfrac{3}{x} = \dfrac{{136}}{{10}}\\ \Rightarrow \dfrac{{13x + 3}}{x} = \dfrac{{136}}{{10}}\\ \Rightarrow 10.\left( {13x + 3} \right) = 136x\\ \Rightarrow 10.13x + 10.3 = 136x\\ \Rightarrow 130x + 30 = 136x\\ \Rightarrow 130x - 136x = - 30\\ \Rightarrow x.\left( {130 - 136} \right) = - 30\\ \Rightarrow \left( { - 6} \right).x = - 30\\ \Rightarrow x = \left( { - 30} \right):\left( { - 6} \right)\\ \Rightarrow x = 5\end{array}\)
Cho các biểu thức:
\(A = \dfrac{{34}}{{7.13}} + \dfrac{{51}}{{13.22}} + \dfrac{{85}}{{22.37}} + \dfrac{{68}}{{37.49}};\)
\( B = \dfrac{{39}}{{7.16}} + \dfrac{{65}}{{16.31}} + \dfrac{{52}}{{31.43}} + \dfrac{{26}}{{43.49}}\)
Tỉ số \(\dfrac{A}{B}\) bằng:
\(A = \dfrac{{34}}{{7.13}} + \dfrac{{51}}{{13.22}} + \dfrac{{85}}{{22.37}} + \dfrac{{68}}{{37.49}}\)
\(= \dfrac{{34}}{6}.\dfrac{6}{{7.13}} + \dfrac{{51}}{9}.\dfrac{9}{{13.22}}\)\(\, + \dfrac{{85}}{{15}}.\dfrac{1}{{22.37}} + \dfrac{{68}}{{12}}.\dfrac{{12}}{{37.49}}\)
\(= \dfrac{{17}}{3}.\left( {\dfrac{1}{7} - \dfrac{1}{{13}}} \right) + \dfrac{{17}}{3}.\left( {\dfrac{1}{{13}} - \dfrac{1}{{22}}} \right) \)\(\,+ \dfrac{{17}}{3}.\left( {\dfrac{1}{{22}} - \dfrac{1}{{37}}} \right) \)\(\,+ \dfrac{{17}}{3}.\left( {\dfrac{1}{{37}} - \dfrac{1}{{49}}} \right)\)
\(= \dfrac{{17}}{3}.\dfrac{1}{7} - \dfrac{{17}}{3}.\dfrac{1}{{13}} + \dfrac{{17}}{3}.\dfrac{1}{{13}} \)\(\,- \dfrac{{17}}{3}.\dfrac{1}{{22}} + \dfrac{{17}}{3}.\dfrac{1}{{22}} - \dfrac{{17}}{3}.\dfrac{1}{{37}} \)\(\,+ \dfrac{{17}}{3}.\dfrac{1}{{37}} - \dfrac{{17}}{3}.\dfrac{1}{{49}}\)
\(= \dfrac{{17}}{3}.\dfrac{1}{7} - \dfrac{{17}}{3}.\dfrac{1}{{49}} = \dfrac{{17}}{3}\left( {\dfrac{1}{7} - \dfrac{1}{{49}}} \right)\)
\(= \dfrac{{17}}{3}.\left( {\dfrac{7}{{49}} - \dfrac{1}{{49}}} \right) = \dfrac{{17}}{3}.\dfrac{6}{{49}} = \dfrac{{34}}{{49}}\)
\(B = \dfrac{{39}}{{7.16}} + \dfrac{{65}}{{16.31}} + \dfrac{{52}}{{31.43}} + \dfrac{{26}}{{43.49}}\)
\(= \dfrac{{39}}{9}.\dfrac{9}{{7.16}} + \dfrac{{65}}{{15}}.\dfrac{{15}}{{16.31}} \)\(\,+ \dfrac{{52}}{{12}}.\dfrac{{12}}{{31.43}} + \dfrac{{26}}{6}.\dfrac{6}{{43.49}}\)
\(= \dfrac{{13}}{3}.\left( {\dfrac{1}{7} - \dfrac{1}{{16}}} \right) + \dfrac{{13}}{3}.\left( {\dfrac{1}{{16}} - \dfrac{1}{{31}}} \right) \)\(\,+ \dfrac{{13}}{3}.\left( {\dfrac{1}{{31}} - \dfrac{1}{{43}}} \right) \)\(\,+ \dfrac{{13}}{3}.\left( {\dfrac{1}{{43}} - \dfrac{1}{{49}}} \right)\)
\(= \dfrac{{13}}{3}.\dfrac{1}{7} - \dfrac{{13}}{3}.\dfrac{1}{{16}} + \dfrac{{13}}{3}.\dfrac{1}{{16}} \)\(\,- \dfrac{{13}}{3}.\dfrac{1}{{31}} + \dfrac{{13}}{3}.\dfrac{1}{{31}} - \dfrac{{13}}{3}.\dfrac{1}{{43}}\)\(\, + \dfrac{{13}}{3}.\dfrac{1}{{43}} - \dfrac{{13}}{3}.\dfrac{1}{{49}}\)
\(= \dfrac{{13}}{3}.\dfrac{1}{7} - \dfrac{{13}}{3}.\dfrac{1}{{49}}\)\(\, = \dfrac{{13}}{3}.\left( {\dfrac{1}{7} - \dfrac{1}{{49}}} \right)\)
\(= \dfrac{{13}}{3}.\left( {\dfrac{7}{{49}} - \dfrac{1}{{49}}} \right) = \dfrac{{13}}{3}.\dfrac{6}{{49}} = \dfrac{{26}}{{49}}\)
\( \Rightarrow \dfrac{A}{B} = \dfrac{{34}}{{49}}:\dfrac{{26}}{{49}} = \dfrac{{34}}{{49}}.\dfrac{{49}}{{26}} = \dfrac{{17}}{{13}}\)
