1/ = 160/27 - 52/27 = 4
2/ => 3/7x + 1/7 = 1 => 3/7x = 6/7 => x = 2
1, \(\frac{40}{9}.\frac{4}{3}-\frac{4}{9}.\frac{13}{3}\)
\(=\frac{4}{9}.10.\frac{4}{3}-\frac{4}{9}.\frac{13}{3}\)
\(=\frac{4}{9}.\left(10.\frac{4}{3}-\frac{13}{3}\right)=\frac{4}{9}.\left(\frac{40}{3}-\frac{13}{3}\right)\)
\(=\frac{4}{9}.9=4\)
2, \(\left[\frac{3}{7}.x+\frac{1}{7}\right]:\frac{1}{4}=4\)
\(\left[\frac{3}{7}x+\frac{1}{7}\right]=4.\frac{1}{4}=1\)
\(\frac{3}{7}x+\frac{1}{7}=1\)
\(\frac{3}{7}x=1-\frac{1}{7}=\frac{6}{7}\)
\(x=\frac{6}{7}:\frac{3}{7}=\frac{6}{7}.\frac{7}{3}=2\)
