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\(\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{10}{7}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}=\frac{\left(\frac{4293}{324}-\frac{708}{324}-\frac{3510}{324}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{30}{21}+\frac{70}{21}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}=\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:\left(-\frac{41}{21}\right)}\)=\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}=100:\left(-\frac{100}{4}\right)=-4\)
\(30+\frac{14}{5}:\left(\frac{24}{150}-\frac{270}{150}-\frac{25}{150}\right)=30+\frac{14}{5}:\left(-\frac{271}{150}\right)=30+\left(-\frac{420}{271}\right)=\frac{7710}{271}\)
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:
\(A=13-2-10+\frac{1}{4}-\frac{5}{27}-\frac{5}{6}=1+\frac{27-20-90}{108}=1-\frac{83}{108}=\frac{25}{108}\)
\(A.230\frac{1}{5}=\frac{25}{108}.230\frac{1}{25}=\frac{25}{108}.\frac{5751}{25}=\frac{5751}{108}=\frac{213}{4}\)
\(A.230\frac{1}{25}+46\frac{3}{4}=\frac{213}{4}+\frac{187}{4}=100\)
\(B=1+\frac{3}{7}+3+\frac{1}{3}=4+\frac{3}{7}+\frac{1}{3}=4+\frac{16}{21}=\frac{100}{21}\)
\(C=12\frac{1}{3}-14\frac{2}{7}=12-14+\frac{1}{3}-\frac{2}{7}=-2+\frac{7-6}{21}=-\frac{41}{21}\)
\(B:C=\frac{100}{21}:\left(-\frac{41}{21}\right)=-\frac{100}{41}\)
\(E=\frac{A.230\frac{1}{25}+46\frac{3}{4}}{B:C}=\frac{100}{-\frac{100}{41}}=-41\)