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\(a,\frac{1}{2}\times\frac{4}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\times\frac{10}{6}\)
\(=\frac{2}{3}\)
\(b,\frac{24}{35}\div\left(\frac{4}{5}\times\frac{8}{7}\right)\)
\(=\frac{24}{35}\div\frac{32}{35}\)
\(=\frac{24}{35}\times\frac{35}{32}\)
\(=\frac{3}{4}\)
\(\frac{6}{5}\)x 5 - \(\frac{17}{5}\)= \(\frac{30}{5}\)- \(\frac{17}{5}\)=\(\frac{13}{5}\)
ta có:4/5*5+4\10*5)-17/5
=20/5+20/10)-17/5
=50/10-17/5
=10/1+(-17)/5
=-33/5
mk nghĩ là vậy !nếu mk làm sai thìcho mk xin lỗi ! đúng thì tích cho mk!nếu k hiểu hỏi lại nha
\(\frac{5}{8}:x=\frac{3}{4}\)
\(x=\frac{5}{8}:\frac{3}{4}\)
\(x=\frac{5}{6}\)
\(\frac{2}{4}+\frac{3}{4}:\frac{1}{2}\)
\(=\frac{2}{4}+\frac{6}{4}\)
\(\frac{8}{4}=2\)
Hok tốt
Ta có :
\(5-\frac{7}{5}:\frac{3}{10}+\frac{2}{5}\)
\(=5-\frac{7}{5}x\frac{10}{3}+\frac{2}{5}\)
\(=5-\frac{14}{3}+\frac{2}{5}\)
\(=\frac{15}{3}-\frac{14}{3}+\frac{6}{15}\)
\(=\frac{1}{3}+\frac{6}{15}\)
\(=\frac{5}{15}+\frac{6}{15}\)
\(=\frac{11}{15}\)
Tham khảo nha !!!!
5 - \(\frac{7}{5}\): \(\frac{3}{10}\)+ \(\frac{2}{5}\)= 5 - \(\frac{14}{3}\)+ \(\frac{2}{5}\)
= \(\frac{1}{3}\)+ \(\frac{2}{5}\)
= \(\frac{11}{15}\)
TL
A \(\frac{1}{4}+\frac{1}{7}=\frac{1\times7}{4\times7}+\frac{1\times4}{7\times4}=\frac{7}{24}+\frac{4}{28}=\frac{11}{28}\)
B\(\frac{1}{3}+\frac{1}{6}=\frac{1\times2}{3\times2}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
C\(\frac{1}{4}+\frac{1}{5}=\frac{1\times5}{4\times5}+\frac{1\times4}{5\times4}=\frac{5}{20}+\frac{4}{20}=\frac{9}{20}\)
\(\frac{11}{28},\frac{1}{2},\frac{9}{20}\)MSC 140
\(\frac{11}{28}=\frac{11\times5}{28\times5}=\frac{55}{140}\)
\(\frac{1}{2}=\frac{1\times70}{2\times70}=\frac{70}{140}\)
\(\frac{9}{20}=\frac{9\times7}{20\times7}=\frac{63}{140}\)
\(\frac{55}{140}< \frac{63}{140}< \frac{70}{140}\)hoặc \(\frac{1}{4}+\frac{1}{7}< \frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{1}{6}\)
Vậy cho B
HT
\(\frac{7}{3}+\frac{4}{3}:\frac{5}{4}\\ =\frac{7}{3}+\frac{4}{3}\cdot\frac{4}{5}\\ =\frac{7}{3}+\frac{16}{15}\\ =\frac{35}{15}+\frac{16}{15}\\ =\frac{51}{15}\\ \frac{17}{15}\)
\(=\frac{7}{3}+\frac{5}{3}\)
\(=\frac{12}{3}=4\)