\(2\frac{2}{7}-\frac{3}{7}+5\frac{2}{7}-\frac{4}{18}-4\cdot\frac{1}{9}+2\frac{9}{21}\)
Tính bằng cách thuận tiện
Mình mong ai đó giúp mình giải bài này
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222225444444455555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555
a \(\frac{4}{18}+\frac{5}{13}+\frac{7}{9}+\frac{21}{13}\)
= \(\left(\frac{4}{18}+\frac{7}{9}\right)+\left(\frac{5}{13}+\frac{21}{13}\right)\)
= \(1+2\)
= \(3\)
b \(\frac{4}{3}-\frac{2}{7}-\frac{5}{7}+\frac{2}{3}\)
= \(\left(\frac{4}{3}-\frac{2}{3}\right)-\left(\frac{2}{7}+\frac{5}{7}\right)\)
= \(\frac{2}{3}-1\)
= \(-\frac{1}{3}\)
a) \(\frac{3}{5}+\frac{4}{7}+\frac{2}{5}+\frac{1}{7}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{4}{7}+\frac{1}{7}+\frac{2}{7}\right)\)
\(=\frac{5}{5}+\frac{7}{7}=1+1=2\)
a) \(\frac{3}{5}+\frac{4}{7}+\frac{2}{5}+\frac{1}{7}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{4}{7}+\frac{1}{7}+\frac{2}{7}\right)\)
= 1 + 1
= 2
b) \(\frac{4}{9}+\frac{8}{9}+\frac{12}{9}+\frac{16}{9}+...+\frac{48}{9}+\frac{52}{9}+\frac{56}{9}\)
\(=\frac{4+8+12+16+...+48+52+56}{9}\)
Xét 4 + 8 + 12 + 16 + ... + 48 + 52 + 56
Số các số hạng là:
(56 - 4) : 4 + 1 = 14 (số)
4 + 8 + 12 + 16 + ... + 48 + 52 + 56 = (56 + 4) x 14 : 2 = 420
\(\frac{4+8+12+16+...+48+52+56}{9}=\frac{420}{9}=\frac{140}{3}\)
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, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)