Tìm \[x\], biết:  

a) \(\,2.{x}\, - \,\frac{5}{4}\, = \,\frac{{20}}{{15}}\)

b) \({\left( { {x} + \frac{1}{5}} \right)^3} = \left( {\frac{{ - 1}}{8}} \right)\)       

c) \(\,\frac{{ - 2}}{3}\,x + 1,6\, = \,\frac{3}{2}\)

d) \(\frac{{243}}{{{3^{x + 1}}}} = {3^x}\)

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