1. Cho hình vẽ bên.

Biết \[d\parallel {\rm{ }}d'\].Tính các góc \[{D_1};\,\,{C_2};{\rm{ }}{C_3};\,\,{B_4}\].

2. Cho hình vẽ:

Biết \[Bx\parallel Cy\] và \(\widehat {ABx} = 130^\circ ,\,\,\widehat {ACy} = 140^\circ .\)

Chứng minh rằng \[AB\] vuông góc với \[AC\].

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

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

 a) \(\frac{7}{8}\,\,.\,\,\left( {\frac{2}{{12}}\, + \,\frac{4}{{10}}} \right) = \frac{7}{8}\left( {\frac{{10}}{{60}} + \frac{{24}}{{60}}} \right)\)

 \[ = \frac{7}{8}.\frac{{17}}{{30}}\]=\(\frac{{119}}{{240}}\)

 b) \(1\frac{3}{7}:\left( { - \frac{4}{5}} \right) - 3\frac{3}{7}:\left( { - \frac{4}{5}} \right) = \left( {\frac{{10}}{7} - \frac{{24}}{7}} \right):\left( { - \frac{4}{5}} \right)\)                            

 \( = \left( {\frac{{10}}{7} - \frac{{24}}{7}} \right).\frac{{ - 5}}{4}\)\[ = \frac{{ - 5}}{4}.( - 2) = \frac{5}{2}\]  

 c) \(\frac{{{9^2}\,.\,{{27}^3}}}{{{{81}^4}}} = \frac{{{{({3^2})}^2}\,.\,{{({3^3})}^3}}}{{{{({3^4})}^4}}} = \frac{{{3^4}\,.\,{3^9}}}{{{3^{16}}}}\)

           \[ = \frac{{{3^{13}}}}{{{3^{16}}}} = \frac{1}{{27}}\]

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

\[ = (2,5\,.\,0,4)\,\,.\,\,\left( { - \frac{1}{8}} \right) + \frac{3}{8}\,\,\left( {41\frac{2}{7} - 38\frac{2}{7}} \right)\]

 \( = - \frac{1}{8} + \frac{3}{8}\,.\,3 = - \frac{1}{8} + \frac{9}{8} = 1\)