NM
6
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
26 tháng 8 2020
Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
12 tháng 4 2019
Mình làm câu a,c, thôi , còn bạn làm nốt bài b,d, nhé
\(a,\frac{7\cdot9-14}{3-17}=\frac{63-14}{3-17}=\frac{49}{-14}=\frac{7\cdot7}{(-2)\cdot7}=\frac{7}{-2}=\frac{-7}{2}\)
\(c,\frac{9}{23}\cdot\frac{5}{8}+\frac{9}{23}\cdot\frac{3}{8}-\frac{9}{23}\)
\(=\frac{9}{23}\cdot\left[\frac{5}{8}+\frac{3}{8}-1\right]\)
\(=\frac{9}{23}\cdot\left[1-1\right]=\frac{9}{23}\cdot0=0\)
26 tháng 7 2020
e. \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}\cdot2=-\frac{6}{5}\)
f. \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{5}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{5}=\frac{1}{3}\cdot2-\frac{4}{5}=\frac{2}{3}-\frac{4}{5}=-\frac{2}{15}\)
g. \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}\cdot1+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
h. \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
31 tháng 5 2020
Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
\(\frac{23}{4}+\frac{7}{4}=\frac{30}{4}=\frac{15}{2}\)
\(\frac{9}{5}-\frac{6}{5}=\frac{3}{5}\)
Tk mình mình tk lại 100% ! ^^
=15/2
=3/5