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

\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)

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

\(=4-1\frac{3}{4}\)

\(=2\frac{1}{4}\)

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

\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)

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

\(=4-\frac{7}{4}\)

\(=\frac{16}{4}-\frac{7}{4}\)

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

\(\frac{4}{5}+\frac{2}{35}\times\frac{7}{2}=\frac{4}{5}+\frac{1}{5}=1\)

\(\frac{4}{5}+\left(\frac{1}{5}-\frac{1}{7}\right):\frac{2}{7}\)

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

\(=\frac{5}{5}-\frac{1}{2}\)

\(=1-\frac{1}{2}=\frac{1}{2}\)

\(a,\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\)

\(b,\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)

\(=\frac{1\times2\times3}{2\times3\times4}=\frac{1}{4}\)

\(\frac{5}{16}+\frac{5}{7}+0,4+\frac{1}{6}-\frac{4}{35}\)

= \(\frac{5}{16}+\frac{25}{35}+\frac{2}{5}+\frac{1}{6}-\frac{4}{35}\)

= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{25}{35}-\frac{4}{35}\)

= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{2}{3}\)

= \(\frac{5}{16}+\frac{2}{5}+\frac{5}{6}\)

= \(\frac{5}{16}+\frac{37}{30}\)

= \(\frac{371}{240}\)

 \(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)có 9 p/số

\(=\frac{1}{10}.9=\frac{9}{10}\)

= 1/10 + 1/10 + ...................... + 1/10 có 9 phân số

= 1/10 x 9

= 9/10

ai tk mình mình đang bị âm điểm

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x (1 - 1/6)

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6

= 1/6 (tử và mẫu triệt tiêu cho nhau)