Tính phép tính:              

a) \(\frac{7}{8}\,\,.\,\,\left( {\frac{2}{{12}}\, + \,\frac{4}{{10}}} \right)\)                          b) \(1\frac{3}{7}:\left( { - \frac{4}{5}} \right) - 3\frac{3}{7}:\left( { - \frac{4}{5}} \right)\) 

c) \(\frac{{{9^2}{{.27}^3}}}{{{{81}^4}}}\)                                 d) \[2,5.{\left( { - \frac{1}{2}} \right)^3}.0,4{\rm{   +   41}}\frac{2}{7}.\frac{3}{8} - 38\frac{2}{7}.\frac{3}{8}\]

1. Ta có: \[d'\parallel d''\]

\( \Rightarrow {\widehat D_1} = \widehat A = 61^\circ \) (hai góc so le trong)

\( \Rightarrow \widehat {{C_2}} = \widehat B = 100^\circ \) (hai góc đồng vị)

Ta thấy : \(\widehat {{B_4}} = \widehat {{C_2}} = 100^\circ \) (hai góc so le trong)

Vì \(\widehat {{C_2}} + \widehat {{C_3}} = 180^\circ \)(hai góc kề bù)

\( \Rightarrow 100^\circ + \widehat {{C_3}} = 180^\circ \)\( \Rightarrow \widehat {{C_3}} = 80^\circ \)

Kẻ tia \[Am\parallel Bx \Rightarrow \widehat {ABx} + \widehat {mAB} = 180^\circ \] (hai góc trong cùng phía)

\[ \Rightarrow \widehat {mAB} = 50^\circ \]

Ta có \(\left\{ \begin{array}{l}Am\parallel Bx\\Bx\parallel Cy\end{array} \right. \Rightarrow Am\parallel Cy\) (ba đường thẳng song song)

\[ \Rightarrow \widehat {yCA} + \widehat {CAm} = 180^\circ \] (hai góc trong cùng phía)

\( \Rightarrow \) \[\widehat {mAC} = 40^\circ \]

Ta có \[\widehat {CAB} = \widehat {CAm} + \widehat {BAm} = 90^\circ \]

\( \Rightarrow \)\(BA \bot CA\)\( \Rightarrow \) đpcm

a) \({2^{225}}\, = \,{\left( {{2^3}} \right)^{75}}\, = \,{8^{75}}\)

 \({3^{150\,}} = \,{\left( {{3^2}} \right)^{75}}\, = \,{9^{75}}\)

 Vì \[{8^{75}} < {9^{75}}\].

 Nên \[{2^{225}} < \;{3^{150}}\]

 b) \({2^{12}}\, = \,{\left( {{2^2}} \right)^6}\, = \,{4^6}\)

  \({4^{18}}\, = \,{\left( {{4^3}} \right)^6}\, = \,{64^6}\)