Tìm \[x\], biết:

          1) \[{\left( {\frac{3}{2}} \right)^3}\,.\,x = {\left( {\frac{3}{2}} \right)^5}\]                            2) \[\frac{1}{2} - x = \frac{3}{4}\]

          3) \[{\left( {x - 0,3} \right)^2} = 9\]                             4) \[\frac{{ - 27}}{8}:x = {\left( {\frac{3}{2}} \right)^3}\]

1) \[ - 0,5 + \frac{3}{2} = \frac{{ - 1}}{2} + \frac{3}{2}\]  

                   \[ = 1\]

 2) \({\left( { - 0,25} \right)^4}{.4^4}\) =  (-0,25.4)4  

                          = (-1)4 = 1

 3) \(\frac{2}{9}.\frac{8}{3} - \frac{{11}}{9}.\frac{8}{3} = \frac{8}{3}.\left( {\frac{2}{9} - \frac{{11}}{9}} \right)\)  

                        \( = \frac{8}{3}.( - 1) =  - \frac{8}{3}\)

 4) \(\frac{{14}}{5} + \left( {2,5 - \frac{9}{5}} \right) = \frac{{14}}{5} - \frac{9}{5} + 2,5\)

                             \( = 1 + 2,5\)\( = 3,5\)

 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)