Thực hiện phép tính:

          1) \[ - 0,5 + \frac{3}{2}\]                                  2) \({\left( { - 0,25} \right)^4}.\,{4^4}\)

          3) \(\frac{2}{9}.\frac{8}{3} - \frac{{11}}{9}.\frac{8}{3}\)                               4) \(\frac{{14}}{5} + \left( {2,5 - \frac{9}{5}} \right)\)

 1) Vì \[{\widehat M_2}\]= \[{\widehat A_2}\] (\[ = 58^\circ \])

 Mà \[{\widehat M_2}\]; \[{\widehat A_2}\] ở vị trí so le trong

 \[ \Rightarrow a\parallel b\]

 2) \[{\widehat M_2}\] = \[{\widehat M_4}\] (hai góc đối đỉnh )  

 Vì \[a\parallel b\]

 \[ \Rightarrow \]\[{\widehat M_2}\] = \[{\widehat A_4}\] (2 góc đồng vị )  

 3) Ta có: \({\widehat A_2} = {\widehat A_4} = 58^\circ \)(hai góc đối đỉnh)

 Mà \({\widehat A_2} + {\widehat A_1} = 180^\circ \) (hai góc kề bù)

\[ \Rightarrow \]\({\widehat A_1} = {\rm{ }}180^\circ - {\widehat A_2} = 180^\circ --58^\circ = 122^\circ \)   

 Do \[a\parallel b\] \( \Rightarrow {\widehat M_1} = {\widehat A_1} = 122^\circ \) (hai góc đồng vị)

 Mà \({\widehat A_1} = {\widehat A_3} = 122^\circ \) (hai góc đối đỉnh)

\[A = \frac{5}{{2.1}} + \frac{4}{{1.11}} + \frac{3}{{11.2}} + \frac{1}{{2.15}} + \frac{{13}}{{15.4}}\]

\[\,\,\,\,\, = 7.\left( {\frac{5}{{2.7}} + \frac{4}{{7.11}} + \frac{3}{{11.14}} + \frac{1}{{14.15}} + \frac{{13}}{{15.28}}} \right)\]

 \[ = 7.\left( {\frac{1}{2} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{14}} + \frac{1}{{11}} - \frac{1}{{14}} + \frac{1}{{14}} - \frac{1}{{15}} + \frac{1}{{15}} - \frac{1}{{28}}} \right)\]

 \[ = 7.\left( {\frac{1}{2} - \frac{1}{{28}}} \right)\]

 \[ = 7.\frac{{13}}{{28}} = \frac{{13}}{4}\]