A=\(\frac{-41000}{973}\)

                                                    Giải:         

Đặt \(A=13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6},B=1\frac{3}{7}+\frac{10}{3},C=12\frac{1}{3}-14\frac{2}{7}\)

Ta có:             

\(A=13-2-10+\frac{1}{4}-\frac{5}{27}-\frac{5}{6}=1+\frac{27-20-90}{108}=1-\frac{83}{108}=\frac{25}{108}\)              

\(A.230\frac{1}{5}=\frac{25}{108}.230\frac{1}{25}=\frac{25}{108}.\frac{5751}{25}=\frac{5751}{108}=\frac{213}{4}\)                        

\(A.230\frac{1}{25}+46\frac{3}{4}=\frac{213}{4}+\frac{187}{4}=100\)                     

\(B=1+\frac{3}{7}+3+\frac{1}{3}=4+\frac{3}{7}+\frac{1}{3}=4+\frac{16}{21}=\frac{100}{21}\)                               

\(C=12\frac{1}{3}-14\frac{2}{7}=12-14+\frac{1}{3}-\frac{2}{7}=-2+\frac{7-6}{21}=-\frac{41}{21}\)            

\(B:C=\frac{100}{21}:\left(-\frac{41}{21}\right)=-\frac{100}{41}\)                

\(E=\frac{A.230\frac{1}{25}+46\frac{3}{4}}{B:C}=\frac{100}{-\frac{100}{41}}=-41\)

\(\frac{\frac{25}{108}.\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)=\(\frac{\frac{5755}{108}+\frac{187}{4}}{\frac{-973}{410}}\)=\(\frac{\frac{8531}{84}}{\frac{-973}{410}}\)=-241,0180

\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

\(=\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right)\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\left(-\frac{41}{21}\right)}\)

\(=\frac{\frac{25}{108}\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}\cdot\left(-\frac{21}{41}\right)}=\frac{\frac{2701}{27}}{-\frac{973}{410}}\)

Tính nốt vì số dữ quá , lần sau để số ít thôi

đoạn \(\frac{8531}{\frac{34}{\frac{-973}{410}}}\)là mình nhầm nha
nó phải là\(\frac{\frac{2701}{27}}{\frac{-973}{410}}\)=-237,405