1+2+3+4+5+6+7+8+9+1000000000=1000000045

đi đã rồi kb sau

a) Vì 4 < 5 nên \(\frac{3}{4}>\frac{3}{5}\)

b) Vì 4 < 5 nên \(\frac{4}{7}< \frac{5}{7}\)

c) Ta có :

\(\frac{2}{7}=\frac{2\times5}{7\times5}=\frac{10}{35}\)

\(\frac{4}{5}=\frac{4\times7}{5\times7}=\frac{28}{35}\)

Vì 10 < 28 nên \(\frac{10}{35}< \frac{28}{35}\)hay \(\frac{2}{7}< \frac{4}{5}\)

d) Ta có :

\(\frac{8}{9}=\frac{8\times10}{9\times10}=\frac{80}{90}\)

\(\frac{9}{10}=\frac{9\times9}{10\times9}=\frac{81}{90}\)

Vì 80 < 81 nên \(\frac{80}{90}< \frac{81}{90}\)hay \(\frac{8}{9}< \frac{9}{10}\)

e) Ta có :

\(\frac{10}{9}=\frac{10\times2}{9\times2}=\frac{20}{18}\)

Vì 19 < 20 nên \(\frac{19}{18}< \frac{20}{18}\)hay \(\frac{19}{18}< \frac{10}{9}\)

f) Ta có :

\(\frac{8}{3}=\frac{8\times4}{3\times4}=\frac{32}{12}\)

\(\frac{5}{6}=\frac{5\times2}{6\times2}=\frac{10}{12}\)

Vì 32 > 10 nên \(\frac{32}{12}>\frac{10}{12}\)hay \(\frac{8}{3}>\frac{5}{6}\)

bn nào dài được đầu tiên mà đúng mik sẽ k cho bn đó nhé

kết quả là: 10890777777777777122 nha bạn

~~~@# chúc bạn lun họk tốt ~~@#

Kết quả là

  45

   còn lại tính rất khó

\(\frac{1}{3}-\frac{1}{5}=\frac{5}{3\times5}-\frac{3}{5\times3}=\frac{2}{3\times5}\Rightarrow\frac{2}{3\times5}=\frac{1}{3}-\frac{1}{5}\)\(;\frac{1}{5}-\frac{1}{7}=\frac{7}{5\times7}-\frac{5}{7\times5}=\frac{2}{5\times7}\Rightarrow\frac{2}{5\times7}=\frac{1}{5}-\frac{1}{7}\)

à ta sẽ dùng phân số trung gian thôi !

a)Có:\(\dfrac{5}{7}>\dfrac{2}{7};\dfrac{2}{7}>\dfrac{2}{8}\)

\(\Rightarrow\dfrac{5}{7}>\dfrac{2}{8}\)

b)Có:\(\dfrac{4}{8}>\dfrac{3}{8};\dfrac{3}{8}>\dfrac{3}{9}\)

\(\Rightarrow\dfrac{4}{8}>\dfrac{3}{9}\)

c)Có:\(\dfrac{2}{9}< \dfrac{2}{8};\dfrac{2}{8}< \dfrac{3}{8}\)

\(\Rightarrow\dfrac{2}{9}< \dfrac{3}{8}\)

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

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

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

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

=> \(x+\frac{43}{5}=\frac{37}{4}\)

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

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

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

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

=> \(x-\frac{61}{8}=\frac{3}{2}\)

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

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

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

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

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

=> \(x.\frac{98}{99}=198\)

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

4 7/10 < 6 7/10

3 4/15 <3 11/15

5 1/9 > 2 2/5

2 2/5 > 2 10/15