Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
\(1.a,\frac{5}{7}=\frac{5.9}{7.9}=\frac{45}{63};\frac{4}{9}=\frac{4.7}{9.7}=\frac{28}{63}.\)
\(b,\frac{7}{15},\frac{5}{3}=\frac{5.5}{3.5}=\frac{25}{15}\)
\(c,\frac{11}{12}=\frac{11.4}{12.4}=\frac{44}{48};\frac{7}{48}\)
\(d,\frac{3}{2}=\frac{3.3}{2.3}=\frac{9}{6};\frac{2}{3}=\frac{2.2}{3.2}=\frac{4}{6}\)
\(e,\frac{1}{3}=\frac{1.4}{3.4}=\frac{4}{12};\frac{5}{4}=\frac{5.3}{4.3}=\frac{15}{12};\frac{10}{12}\)
a) $\frac{{14}}{{18}}:\frac{8}{9} = \frac{7}{9}:\frac{8}{9} = \frac{7}{9} \times \frac{9}{8} = \frac{{63}}{{72}} = \frac{7}{8}$
b) $\frac{9}{6}:\frac{3}{{10}} = \frac{3}{2}:\frac{3}{{10}} = \frac{3}{2} \times \frac{{10}}{3} = \frac{{30}}{6} = 5$
c) $\frac{4}{5}:\frac{{10}}{{15}} = \frac{4}{5}:\frac{2}{3} = \frac{4}{5} \times \frac{3}{2} = \frac{{12}}{{10}} = \frac{6}{5}$
d) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
1)\(\frac{7}{8}>\frac{6}{7}>\frac{4}{5}>\frac{1}{2}>\frac{5}{16}\)
2)
a.\(\frac{3}{7}\)và\(\frac{5}{16}\)
Ta có :\(\frac{3}{7}=\frac{3\times5}{7\times5}=\frac{15}{35}\) \(\frac{5}{16}=\frac{5\times3}{16\times3}=\frac{15}{48}\)
\(\frac{15}{35}>\frac{15}{48}\Rightarrow\frac{3}{7}>\frac{5}{16}\)
b.làm tương tự như câu a nhé
Câu 1:
B
1B
đăng từng bài nha