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\(B=\frac{2,5-4.\left(\frac{5}{2}-1,2\right)+\frac{3}{8}}{4.\left(\frac{5}{2}-1,2\right)-\frac{3}{5}:\frac{2}{5}}-\frac{55}{148}\)
\(B=\frac{\frac{5}{2}-4.\left(\frac{25}{10}-\frac{12}{10}\right)+\frac{3}{8}}{4.\left(\frac{25}{10}-\frac{12}{10}\right)-\frac{3}{5}.\frac{5}{2}}-\frac{55}{148}\)
\(B=\frac{\frac{5}{2}-4.\frac{13}{10}+\frac{3}{8}}{4.\frac{13}{10}-\frac{3}{2}}-\frac{55}{148}\)
\(B=\frac{\frac{5}{2}-\frac{26}{5}+\frac{3}{8}}{\frac{26}{5}-\frac{3}{2}}-\frac{55}{148}\)
\(B=\frac{\frac{100}{40}-\frac{208}{40}+\frac{15}{40}}{\frac{52}{10}-\frac{15}{10}}-\frac{55}{148}\)
\(B=\frac{-\frac{93}{40}}{\frac{37}{10}}-\frac{55}{148}\)
\(B=\frac{93}{148}-\frac{55}{148}\)
\(B=\frac{19}{74}\)
\(B=\frac{2,5-4.\left(\frac{5}{2}-1,2\right)+\frac{3}{8}}{4.\left(\frac{5}{2}-1,2\right)-\frac{3}{5}:\frac{2}{5}}-\frac{55}{148}\)
\(B=\frac{\frac{5}{2}-4.\left(\frac{25}{10}-\frac{12}{10}\right)+\frac{3}{8}}{4.\left(\frac{25}{10}-\frac{12}{10}\right)-\frac{3}{5}.\frac{5}{2}}-\frac{55}{148}\)
\(B=\frac{\frac{5}{2}-4.\frac{13}{10}+\frac{3}{8}}{4.\frac{13}{10}-\frac{3}{2}}-\frac{55}{148}\)
\(B=\frac{\frac{5}{2}-\frac{26}{5}+\frac{3}{8}}{\frac{26}{5}-\frac{3}{2}}-\frac{55}{148}\)
\(B=\frac{\frac{100}{40}-\frac{208}{40}+\frac{15}{40}}{\frac{52}{10}-\frac{15}{10}}-\frac{55}{148}\)
\(B=\frac{-\frac{93}{40}}{\frac{37}{10}}-\frac{55}{148}\)
\(B=\frac{93}{148}-\frac{55}{148}\)
\(B=\frac{19}{74}\)
\(E=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(E=\frac{\frac{4}{5}:\frac{4}{5}:1,25}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{36}{17}}+\frac{6}{5}.\frac{1}{2}.\frac{5}{4}\)
\(E=\frac{1:\frac{5}{4}}{\frac{3}{5}}+\frac{1.\frac{7}{4}}{\frac{119}{36}.\frac{36}{17}}+\frac{3}{4}\)
\(E=\frac{4}{5}.\frac{5}{3}+\frac{\frac{7}{4}}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{7}{4}.\frac{1}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
\(E=\frac{4}{3}+1=\frac{7}{3}\)
Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
\(A=\frac{2,5-4.\left(\frac{5}{2}-1,2\right)+\frac{3}{8}}{4.\left(\frac{5}{2}-1,2\right)-\frac{3}{5}:\frac{2}{5}}-\frac{55}{148}\)
\(A=\frac{\frac{5}{2}-4.\left(\frac{25}{10}-\frac{12}{10}\right)+\frac{3}{8}}{4.\left(\frac{25}{10}-\frac{12}{10}\right)-\frac{3}{5}.\frac{5}{2}}-\frac{55}{148}\)
\(A=\frac{\frac{5}{2}-4.\frac{13}{10}+\frac{3}{8}}{4.\frac{13}{10}-\frac{3}{2}}-\frac{55}{148}\)
\(A=\frac{\frac{5}{2}-\frac{26}{5}+\frac{3}{8}}{\frac{26}{5}-\frac{3}{2}}-\frac{55}{148}\)
\(A=\frac{\frac{100}{40}-\frac{208}{40}+\frac{15}{40}}{\frac{52}{10}-\frac{15}{10}}-\frac{55}{148}\)
\(A=\frac{-\frac{93}{40}}{\frac{37}{10}}-\frac{55}{148}\)
\(A=\frac{93}{148}-\frac{55}{148}\)
\(A=\frac{19}{74}\)