Giải bài 2 (6.28) trang 18 vở thực hành Toán 6

Đề bài

Tính:

a) \(\frac{7}{8} + \frac{7}{8}:\frac{1}{8} - \frac{1}{2}\)                  

b) \(\frac{6}{{11}} + \frac{{11}}{3}.\frac{3}{{22}}\)

Lời giải chi tiết

a) \(\begin{array}{l}\frac{7}{8} + \frac{7}{8}:\frac{1}{8} - \frac{1}{2} = \frac{7}{8} + \frac{7}{8}.\frac{8}{1} - \frac{1}{2} = \frac{7}{8} + 7 - \frac{1}{2} = \left( {\frac{7}{8} - \frac{1}{2}} \right) + 7\\ = \left( {\frac{7}{8} - \frac{4}{8}} \right) + 7 = \frac{3}{8} + 7 = 7\frac{3}{8}\end{array}\)

b) \(\frac{6}{{11}} + \frac{{11}}{3}.\frac{3}{{22}} = \frac{6}{{11}} + \frac{{11}}{{22}} = \frac{{12}}{{22}} + \frac{{11}}{{22}} = \frac{{23}}{{22}}\)