Giải bài 1 (6.38) trang 22 vở thực hành Toán 6

Đề bài

Bài 1 (6.38). Tính:

a) \(\frac{{ - 1}}{2} + \frac{5}{6} + \frac{1}{3};\)                   

b) \(\frac{{ - 3}}{8} + \frac{7}{4} - \frac{1}{{12}};\)

c) \(\frac{3}{5}:\left( {\frac{1}{4}.\frac{7}{5}} \right);\)                   

d) \(\frac{{10}}{{11}} + \frac{4}{{11}}:4 - \frac{1}{8}.\)

Lời giải chi tiết

a) \(\frac{{ - 1}}{2} + \frac{5}{6} + \frac{1}{3} = \frac{{ - 3}}{6} + \frac{5}{6} + \frac{2}{6} = \frac{4}{6} = \frac{2}{3}\)           

b) \(\frac{{ - 3}}{8} + \frac{7}{4} - \frac{1}{{12}} = \frac{{ - 9}}{{24}} + \frac{{28}}{{24}} - \frac{2}{{24}} = \frac{{17}}{{24}}\)

c) \(\frac{3}{5}:\left( {\frac{1}{4}.\frac{7}{5}} \right) = \frac{3}{5}:\frac{7}{{20}} = \frac{3}{5}.\frac{{20}}{7} = \frac{{12}}{7}\)             

d) \(\frac{{10}}{{11}} + \frac{4}{{11}}:4 - \frac{1}{8} = \frac{{10}}{{11}} + \frac{4}{{11}}.\frac{1}{4} - \frac{1}{8} = \frac{{10}}{{11}} + \frac{1}{{11}} - \frac{1}{8} = \frac{{11}}{{11}} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}\)