Question

[Takitori]

Tính

a) \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)

b) \(\frac{298}{719}:\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)

Answer 1

[Kiệt Nguyễn]

\(\frac{2}{3}+\frac{1}{3}\left(\frac{-4}{9}+\frac{5}{6}\right)\div\frac{7}{12}\)

\(=\frac{2}{3}+\frac{1}{3}\left(\frac{-4}{9}+\frac{5}{6}\right)\times\frac{12}{7}\)

\(=\frac{2}{3}+\frac{4}{7}\left(\frac{-4}{9}+\frac{5}{6}\right)\)

\(=\frac{2}{3}+\frac{4}{7}\left(\frac{-8}{18}+\frac{15}{18}\right)\)

\(=\frac{2}{3}+\frac{4}{7}.\frac{-7}{18}\)

\(=\frac{6}{9}+\frac{-2}{9}\)

\(=\frac{4}{9}\)

Answer 2

[Kiệt Nguyễn]

\(\frac{298}{719}\div\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)

\(=\frac{298}{719}\div\left(\frac{3}{12}+\frac{1}{12}-\frac{4}{12}\right)-\frac{2011}{2012}\)

\(=\frac{298}{719}\div0-\frac{2011}{2012}\)

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