Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ 27\\ b.\ \frac{14}{3} \end{array}$Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ \left[\frac{3}{7} .\frac{4}{15} +\frac{1}{3} .\left( 9^{15}\right)\right]^{0} .\frac{1}{3} .\frac{6^{8}}{12^{4}}\\ =1.\frac{1}{3} .\frac{2^{8} .3^{8}}{2^{8} .3^{4}} =\frac{3^{8}}{3^{5}} =3^{3} =27\\ b.\ \frac{10^{4} .81-16.15^{2}}{4^{4} .675} =\frac{2^{4} .5^{4} .3^{4} -2^{4} .3^{2} .5^{2}}{2^{8} .5^{2} .3^{3}}\\ =\frac{2^{4} .3^{2} .5^{2} .\left( 5^{2} .3^{2} -1\right)}{2^{8} .5^{2} .3^{3}} =\frac{225-1}{2^{4} .3} =\frac{224}{48} =\frac{14}{3}\\ Bài\ 2:\\ Ta\ có:\ \frac{a}{b} =\frac{c}{d} \Leftrightarrow ad=bc\\ Giả\ sử\ \frac{a^{2} +ac}{c^{2} -ac} =\frac{b^{2} +bd}{d^{2} -bd}\\ \Leftrightarrow \left( a^{2} +ac\right) .\left( d^{2} -bd\right) =\left( c^{2} -ac\right) .\left( b^{2} +bd\right)\\ \Leftrightarrow a^{2} d^{2} +acd^{2} -a^{2} bd-abcd=b^{2} c^{2} +bdc^{2} -b^{2} ac-abcd\\ \Leftrightarrow ad( ad+cd-ab) =bc( bc+dc-ba) \ mà\ ad=bc\\ \Leftrightarrow cd-ab=cd-ab( luôn\ đúng)\\ \Rightarrow Giả\ sử\ đúng\Rightarrow đpcm \end{array}$
