\(\frac{15}{25}=\frac{3\cdot5}{5\cdot5}=\frac{3}{5}=\frac{21}{35}< \frac{25}{35}=\frac{5\cdot5}{5\cdot7}=\frac{5}{7}\)

\(\frac{17}{49}< \frac{17}{48}< \frac{18}{48}=\frac{3\cdot6}{6\cdot8}=\frac{3}{8}\)

\(\frac{13}{60}< \frac{13}{50}=\frac{26}{100}< \frac{27}{100}\)

\(\frac{43}{47}>\frac{42}{47}>\frac{42}{49}=\frac{6\cdot7}{7\cdot7}=\frac{6}{7}=\frac{30}{35}>\frac{29}{35}\)

\(\frac{1993}{1995}< \frac{1994}{1996}=\frac{997}{998}\)

Đáp án là 24

Câu trả lời là :8/7=72/63.         5/9=35/63.         4/9=16/36.             7/12=21/36.          1/2=6/12.          5/12=5/12.            2/3=8/12

Chúc cậu học tốt, nha

\(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right)\div16\frac{2}{3}=0\)

\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right)\div\frac{50}{3}=0\)

\(\left(\frac{43}{8}+x-\frac{173}{24}\right)\div\frac{50}{3}=1-0\)

\(\frac{43}{8}+x-\frac{173}{24}=1\times\frac{50}{3}=\frac{50}{3}\)

\(\frac{43}{8}+x=\frac{50}{3}+\frac{173}{24}=\frac{191}{8}\)

\(\Rightarrow x=\frac{191}{8}-\frac{43}{8}=\frac{148}{8}=\frac{37}{2}\)

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)

- Vc Lớp 5 

32,2857 và 12,8 và -498814315,3 và 0,1 và 11 

tổng = - 498814259,1