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c )
\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{5}{3}}}=1+\frac{1}{1+\frac{1}{\frac{8}{3}}}=1+\frac{1}{\frac{11}{8}}=\frac{19}{11}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)
\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)
\(A=1-1=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)
\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)
\(A=0-\frac{43}{101}=\frac{-43}{101}\)
\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)
\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)
\(C=\frac{-2}{1}=-2\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)
\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)
b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)
c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)
\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)
d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)
Tính:
a) \(-3\frac{3}{4}+\left(-\frac{10}{25}\right)+\left(-\frac{6}{12}\right)\)
\(=\left(-\frac{15}{4}\right)+\left(-\frac{10}{25}\right)+\left(-\frac{6}{12}\right)\)
\(=\left(-\frac{83}{20}\right)+\left(-\frac{6}{12}\right)\)
\(=-\frac{93}{20}.\)
b) \(-0,6-\left(-\frac{4}{9}\right)-\frac{16}{15}\)
\(=\left(-\frac{3}{5}\right)-\left(-\frac{4}{9}\right)-\frac{16}{15}\)
\(=\left(-\frac{7}{45}\right)-\frac{16}{15}\)
\(=-\frac{11}{9}.\)
Chúc bạn học tốt!
Nhiều thế :( Làm 1,2 câu thôi nhé
a) \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\) (bị mất nét nhưng vẫn nhìn ra là số 12 nhỉ?)
b) \(\frac{-2}{5}+\frac{7}{21}=\frac{-42}{105}+\frac{35}{105}=\frac{-7}{105}=\frac{-1}{15}\)
\(-2\frac{1}{15}-\left|\frac{-5}{12}\right|+1\frac{5}{12}\)\(+\left|\frac{-31}{15}\right|\)
\(=\frac{-31}{15}-\frac{5}{12}+\frac{17}{12}+\frac{31}{15}\)
\(=\frac{-31}{15}+\frac{31}{15}-\frac{5}{12}+\frac{17}{12}\)
\(=0+1=1\)