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bài 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)

\(\frac37+\frac{5}{13}+\frac{4}{13}\)
=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)
=\(\frac37+\frac{9}{13}\)
=\(\frac{39}{91}+\frac{63}{91}\)
=\(\frac{102}{91}\)
\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)

1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)

a. \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5=\frac{8}{27}-\frac{9}{16}.\left(-1\right)=\frac{8}{27}+\frac{9}{16}=\frac{371}{432}\)
b. \(12:\left(\frac{3}{4}-\frac{5}{6}\right)^2=12:\left(-\frac{1}{12}\right)^2=12:\frac{1}{144}=12.144=1728\)
c. \(\frac{7}{22}:\frac{3}{11}+\frac{7}{22}:\frac{4}{11}=\frac{7}{22}.\frac{11}{3}+\frac{7}{22}.\frac{11}{4}=\frac{7}{22}\left(\frac{11}{3}+\frac{11}{4}\right)\)
\(=\frac{7}{22}.\frac{77}{12}=\frac{49}{24}\)
d. \(\frac{12}{35}\left(\frac{7}{4}+\frac{13}{4}\right)-\frac{1}{3}=\frac{12}{35}.5-\frac{1}{3}=\frac{12}{7}-\frac{1}{3}=\frac{29}{21}\)

a)\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{14}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)
b)\(\frac12+\frac34-\left(\frac34-\frac45\right)\)
=\(\frac12+\frac34-\frac34+\frac45\)
=\(\left(\frac12+\frac45\right)+\left(\frac34-\frac34\right)\)
=\(\left(\frac{5}{10}+\frac{8}{10}\right)+0\)
=\(\frac{13}{10}\)

a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)