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\(A=\frac{1}{3}.\frac{-9}{10}.\frac{-6}{13}.\frac{-13}{36}=\frac{-3}{10}.\frac{-1}{6}=\frac{1}{20}\)
\(B=\frac{4}{19}\left(\frac{-5}{12}+\frac{-7}{12}\right)-\frac{40}{57}=\frac{-4}{19}-\frac{40}{57}=\frac{-52}{57}\)
2 câu còn lại tự làm
\(A=\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(A=\frac{1}{1}.\frac{-2}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(A=\frac{-2}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(A=\frac{-1}{13}.\frac{-9}{5}.\frac{-13}{36}\)
\(A=\frac{-1}{13}.\frac{-1}{5}.\frac{-13}{4}\)
\(A=\frac{-13}{260}=\frac{-1}{20}\)
Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
Trả lời
\(A=\frac{7}{12}+\frac{5}{12}\div6-\frac{11}{36}\)
\(A=\frac{7}{12}+\frac{5}{12}\times\frac{1}{6}-\frac{11}{36}\)
\(A=\frac{7}{12}+\frac{5}{72}-\frac{11}{36}\)
\(A=\frac{25}{72}\)
\(B=12\frac{1}{3}-\frac{5}{7}\div\left(24-23\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\left(24-\frac{166}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\frac{2}{7}\)
\(B=\frac{37}{3}-\frac{5}{7}\times\frac{7}{2}\)
\(B=\frac{37}{3}-\frac{5}{2}\)
\(B=\frac{59}{6}\)
Bài làm:
Ta có:
\(A=\frac{7}{12}+\frac{5}{12}\div6-\frac{11}{36}\)
\(A=\frac{7}{12}+\frac{5}{72}-\frac{11}{36}\)
\(A=\frac{42+5-22}{72}=\frac{25}{72}\)
và
\(B=12\frac{1}{3}-\frac{5}{7}\div\left(24-23\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\left(24-23-\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}\div\left(1-\frac{5}{7}\right)\)
\(B=\frac{37}{3}-\frac{5}{7}+1\)
\(B=\frac{265}{21}\)
a) \(\frac{13}{26}-\frac{1}{3}-\frac{1}{2}+\frac{7}{21}\)
\(=\frac{1}{2}-\frac{1}{3}-\frac{1}{2}+\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)
\(=0+0\)
\(=0\)
b) \(\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\right)\)
\(=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{17}+\frac{5}{12}\)
\(=\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{7}{17}+\frac{5}{17}\right)+\frac{6}{11}\)
\(=0+\frac{12}{17}+\frac{6}{11}\)
\(=\frac{132}{187}+\frac{102}{187}\)
\(=\frac{234}{187}\)
c) \(\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{12}{10}\right)\)
\(=\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{6}{5}\right)\)
\(=\frac{13}{5}+\frac{7}{16}-\frac{11}{16}+\frac{6}{5}\)
\(=\left(\frac{13}{5}+\frac{6}{5}\right)+\left(\frac{7}{16}-\frac{11}{16}\right)\)
\(=\frac{19}{5}+\left(\frac{-4}{16}\right)\)
\(=\frac{19}{5}-\frac{1}{4}\)
\(=\frac{76}{20}-\frac{5}{20}\)
\(=\frac{71}{20}\)
d) \(-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{21}{30}-\frac{5}{11}\right)\)
\(=-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{7}{10}-\frac{5}{11}\right)\)
\(=-\frac{3}{10}+\frac{6}{11}-\frac{7}{10}+\frac{5}{11}\)
\(=
\left(-\frac{3}{10}-\frac{7}{10}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)\)
\(=\frac{-10}{10}+\frac{11}{11}\)
\(=-1+1\)
\(=0\)
\(A=\frac{7}{12}+\frac{5}{12}:6-\frac{1}{36}\)
\(=\frac{7}{12}+\frac{5}{72}-\frac{11}{36}\)
\(=\frac{42}{72}+\frac{5}{72}-\frac{22}{72}\)
\(=\frac{25}{72}\)
Bài 1:
\(g.\frac{5}{11}+\frac{6}{11}=\frac{5+6}{11}=\frac{11}{11}=1\)\(\)
\(e.\frac{-17}{25}.\frac{20}{33}+\frac{-17}{25}.\frac{13}{33}+\frac{-3}{25}=\frac{-17}{25}.\left(\frac{20}{33}+\frac{13}{33}\right)+\frac{-3}{25}\)
\(=\frac{-17}{25}.1+\frac{-3}{25}=\frac{-17}{25}+\frac{-3}{25}=\frac{-17-3}{25}=\frac{-20}{25}=\frac{-4}{5}\)
\(d.\frac{5}{7}.\frac{19}{23}+\frac{5}{7}.\frac{5}{23}-\frac{5}{7}.\frac{1}{23}=\frac{5}{7}\left(\frac{19}{23}+\frac{5}{23}-\frac{1}{23}\right)\)
\(=\frac{5}{7}\left(\frac{19+5-1}{23}\right)=\frac{5}{7}.1=\frac{5}{7}\)
\(c.\left(-11\right).\frac{9}{22}=\frac{\left(-11\right).9}{22}=\frac{-99}{22}=\frac{-9}{2}\)
\(b.\frac{5}{6}-\frac{1}{8}=\frac{5.4}{6.4}-\frac{1.3}{8.3}=\frac{20}{24}-\frac{3}{24}=\frac{17}{24}\)
\(a.\frac{2}{3}+\frac{1}{5}-\frac{1}{6}=\frac{2.10}{3.10}+\frac{1.6}{5.6}-\frac{1.5}{6.5}=\frac{20}{30}+\frac{6}{30}-\frac{5}{30}\)
\(=\frac{20+6-5}{30}=\frac{21}{30}=\frac{7}{10}\)
Bài 2:
\(a.\frac{3}{4}+x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{3}{4}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{9}{12}\)
\(\Leftrightarrow x=\frac{2}{12}=\frac{1}{6}\)
\(b.x-\frac{11}{12}=0,5\)
\(\Leftrightarrow x=\frac{1}{2}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{6}{12}+\frac{11}{12}\)
\(\Leftrightarrow x=\frac{17}{12}\)
*)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}\)
\(=\frac{5}{6}\)
*)\(\frac{2003}{1.2}+\frac{2003}{2.3}+\frac{2003}{3.4}+...+\frac{2003}{2002.2003}\)
\(=\frac{2003}{1}-\frac{2003}{2}+\frac{2003}{2}-\frac{2003}{3}+\frac{2003}{3}-\frac{2003}{4}+...+\frac{2003}{2002}-\frac{2003}{2003}\)
\(=2003-1\)
\(=2002\)
\(\frac{5}{12}\cdot\frac{19}{6}+\frac{1}{6}-\frac{5}{12}\cdot\frac{7}{6}\)
\(=\frac{5}{12}\left(\frac{19}{6}-\frac{7}{6}\right)+\frac{1}{6}\)
\(=\frac{5}{12}\cdot\frac{19-7}{6}+\frac{1}{6}\)
\(=\frac{5}{12}\cdot\frac{12}{6}+\frac{1}{6}\)
\(=\frac{5\cdot12}{12\cdot6}+\frac{1}{6}\)
\(=\frac{5}{6}+\frac{1}{6}\)
\(=\frac{5+1}{6}=\frac{6}{6}=1\)
= 5/12*16/9+2/12-5/12*7/6
= 5/12*(19/6+1-7/6)
= 5/12*3
= 5/4