Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
#)Giải :
a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
= \(\frac{1}{5}-\frac{1}{25}\)
= \(\frac{4}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\)
= \(\frac{100}{101}\)
c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
= \(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
= \(5\frac{2}{7}\)
= \(\frac{37}{7}\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
a,\(\frac{3}{7}.\frac{4}{9}+\frac{3}{7}.\frac{5}{9}+\frac{5}{14}\)
\(=\frac{3}{7}.\left(\frac{4}{9}+\frac{5}{9}\right)+\frac{5}{14}\)
\(=\frac{3}{7}.1+\frac{5}{14}\)
\(=\frac{3}{7}+\frac{5}{14}=\frac{6}{14}+\frac{5}{14}=\frac{11}{14}\)
b,\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{9}.\frac{-11}{23}-\frac{1}{23}\)
\(=\)\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{9}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.\frac{110}{63}-\frac{1}{23}\)
=\(\frac{-1210}{1449}\)-\(\frac{1}{23}\)
\(=\frac{-1273}{1449}\)
a) A = \(\frac{19}{23}.\frac{-4}{27}-\frac{4}{23}.\frac{2}{7}\)
= \(\frac{19}{7}.\frac{-4}{23}+\frac{-4}{23}.\frac{2}{7}\)
= \(\frac{-4}{23}.\left(\frac{19}{7}+\frac{2}{7}\right)\)
= \(\frac{-4}{23}.3\)
= \(\frac{-12}{23}\)
b) B = \(\frac{3}{5}+\frac{2}{5}.\frac{-11}{3}+\frac{2}{3}.\frac{-2}{5}+\frac{14}{15}\)
= \(\frac{9+14}{15}+\frac{2}{5}.\frac{-11}{3}+\frac{-2}{3}.\frac{2}{5}\)
= \(\frac{23}{15}+\frac{2}{5}\left(\frac{-11}{3}+\frac{-2}{3}\right)\)
= \(\frac{23}{15}+\frac{2}{5}.\frac{-13}{3}\)
= \(\frac{23}{15}+\frac{-26}{15}\)
= \(\frac{-3}{15}=\frac{-1}{5}\)
\(\frac{-5}{7}.\frac{9}{11}+\frac{-5}{7}.\frac{2}{11}+\frac{5}{7}\)
\(=\frac{-5}{7}.\left(\frac{9}{11}+\frac{2}{11}\right)+\frac{5}{7}\)
\(=\frac{-5}{7}+\frac{5}{7}\)
\(=0\)
t tôi nhé
\(\frac{-5}{7}.\frac{9}{11}+\frac{-5}{7}.\frac{2}{11}+\frac{5}{7}\)
=\(\frac{-5}{7}.\left(\frac{9}{11}+\frac{2}{11}\right)+\frac{5}{7}\)
=\(\frac{-5}{7}.1+\frac{5}{7}\)
=\(\frac{-5}{7}+\frac{5}{7}\)
=\(0\)