Bài 16 trang 13 SGK Toán 7 tập 1

Đề bài

a) \((\frac{-2}{3} + \frac{3}{7}): \frac{4}{5} + (\frac{-1}{3} + \frac{4}{7}) : \frac{4}{5}\)

b) \(\frac{5}{9}: (\frac{1}{11} - \frac{5}{22}) + \frac{5}{9} :(\frac{1}{15} - \frac{2}{3})\)

Hướng dẫn giải

a)  \((\frac{-2}{3} + \frac{3}{7}): \frac{4}{5} + (\frac{-1}{3} + \frac{4}{7}) : \frac{4}{5}\)

\(=(\frac{-2}{3}+ \frac{3}{7}+ \frac{-1}{3} +\frac{4}{7}):\frac{4}{5} \)

\(= (\frac{-3}{3} +\frac{7}{7}): \frac{4}{5} = (-1+1):\frac{4}{5} = 0\)

b) \(\frac{5}{9}: (\frac{1}{11} - \frac{5}{22}) + \frac{5}{9} :(\frac{1}{15} - \frac{2}{3})\) 

\(=\frac{5}{9}: \frac{2 - 5}{22} + \frac{5}{9}: \frac{1-10}{15} = \frac{5}{9}.\frac{22}{-3}+\frac{5}{9}.\frac{15}{-9}\)

\( = \frac{5}{9}.(\frac{22}{-3} + \frac{15}{-9}) = {5 \over 9}.\left( {{{ - 22} \over 3} + {{ - 5} \over 3}} \right) \)

\(= {5 \over 9}.{{ - 27} \over 3} = {5 \over 9}.( - 9) = {{5.( - 9)} \over 9} =  - 5\)