\(1\frac{13}{15}\)  . 0,75 - ( 11/20 +25% ) :3/7
=\(\frac{28}{15}\) . 3/4 - ( 11/20 +25/100 ) :3/7
=28/15. 3/4- 4/5 : 3/7
=7/5 - 28/15
=-7/15
Thấy đúng thì bấm đúng cho mh nha!!!
 

\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+\frac{25}{100}\right):\frac{3}{7}\)

= \(\frac{7}{5}-\frac{4}{5}:\frac{3}{7}\)

= \(\frac{7}{5}-\frac{28}{15}\)

= \(-\frac{7}{15}\)

\(1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+\frac{25}{100}\right):\frac{3}{7}\)

=>\(\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{55}{100}+\frac{25}{100}\right)\cdot\frac{7}{3}\)

=>\(\frac{84}{60}-\frac{80}{100}\cdot\frac{7}{3}\)

=>\(\frac{7}{5}-\frac{4}{5}\cdot\frac{7}{3}=\frac{7}{5}-\frac{28}{15}\)

=>\(\frac{7}{5}-\frac{18}{14}=\frac{98}{60}-\frac{90}{60}=\frac{8}{60}=\frac{4}{30}\)

hi mình là kudo skibidi

Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé

\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)

\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)

\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)

Thanks "Blue Moon" nhìu nha

a)\(\dfrac{25}{42}-\dfrac{20}{36}=\dfrac{900}{1512}-\dfrac{840}{1512}=\dfrac{60}{1512}=\dfrac{5}{126}\)

b)\(\dfrac{9}{50}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{4050}{22500}-\dfrac{3900}{22500}-\dfrac{3750}{22500}=\dfrac{-3600}{22500}=\dfrac{-4}{25}\)

1. 336 : x = 12

x = 336 : 12

x = 28

2. (x + 7) - 25 = 13

x + 7 = 13 + 25

x + 7 = 38

x = 38 - 7

x = 31

3. 87 - (73 - x) = 20

73 - x = 87 - 20

73 - x = 67

x = 73 - 67

x = 6

4. x - 105 : 21 = 15

x - 5 = 15

x = 15 + 5

x = 20

5. (x - 36) : 18 = 12

x - 36 = 216

x = 216 + 36

x = 252

\(\frac{25}{12}-\frac{20}{63}=\frac{445}{252}\)

\(\frac{9}{50}-\frac{13}{75}=\frac{1}{150}\)

\(\frac{2}{19}-\frac{2}{65}-\frac{4}{39}=\frac{-8}{285}\)

đơn giải thôi nhưng mình ko bấn fx đc

​

xét vế trái : \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{221}\)

ta có : \(T< \frac{1}{5}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\)

             \(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

             \(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

              \(=\frac{1}{5}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)< \frac{1}{5}+\frac{1}{4}\Rightarrow T< \frac{9}{20}\)