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Ta có :
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(A=3.\frac{6}{25}\)
\(A=\frac{18}{25}\)
Vậy \(A=\frac{18}{25}\)
Chúc bạn học tốt ~
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)
\(=\frac{3.4.6}{25.4}\)
\(\Rightarrow A=\frac{18}{25}\)
\(a,\frac{2001}{2002}.\frac{5}{7}.\frac{2002}{5}.\frac{7}{2001}=\left(\frac{2001}{2002}.\frac{7}{2001}\right).\left(\frac{5}{7}.\frac{2002}{5}\right)\)
\(=\frac{7}{2002}.\frac{2002}{7}=1\)
\(b,\frac{5}{7}.\frac{7}{9}.\frac{9}{11}.\frac{11}{13}=\left(\frac{5}{7}.\frac{7}{9}\right).\left(\frac{9}{11}.\frac{11}{13}\right)=\frac{5}{9}.\frac{9}{13}\)
\(=\frac{5}{13}\)
\(\frac{2000}{2001}=1-\frac{1}{2001}<1-\frac{1}{2002}=\frac{2001}{2002}\)
1 - 2000/2001 = 1/2001 ; 1 - 2001/2002 = 1/2002
2000/2001 < 2001/2002
\(\frac{2}{3}:\frac{5}{6}=\frac{4}{5}\)
\(\frac{4}{5}\cdot\frac{2}{3}=\frac{8}{15}\)
Vậy : \(\frac{2}{3}:\frac{5}{6}< \frac{4}{5}x\cdot\frac{2}{3}\)nên \(\frac{4}{5}< \frac{8}{15}\)
Mình làm câu A
\(\frac{1}{2}\cdot\frac{3}{4}=\frac{3}{8}\)\(;\frac{5}{8}:\frac{6}{8}=\frac{5}{6}\)
Vậy \(\frac{1}{2}\cdot\frac{3}{4}< \frac{5}{8}:\frac{6}{8}\)
Nên : \(\frac{3}{8}< \frac{5}{6}\)
5/6 < 5/4
7/3 > 7/5
2001/2002 < 2001/2000
222/555 < 444/333