\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-...-\dfrac{1}{6}-\dfrac{1}{2}=-\left(-\dfrac{9}{10}+\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+...+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=-\left(-\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=-\left(-\dfrac{9}{10}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{10}\right)=-\left(-\dfrac{9}{10}+\dfrac{9}{10}\right)=0\)

Ta có :

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{9.10}+\dfrac{1}{9.8}+\dfrac{1}{7.6}+\dfrac{1}{6.5}+\dfrac{1}{5.4}+\dfrac{1}{4.3}+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+..........+\dfrac{1}{2}-\dfrac{1}{1}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{9}{10}-\dfrac{9}{10}=0\)

Đầu tiên , ta cộng các phần nguyên lại với nhau trước :

 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{8}{72}+\frac{1}{90}+\frac{1}{10}\)

= 45 + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{42}+\frac{1}{72}\right)+\left(\frac{1}{10}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{90}\right)+\frac{1}{56}\)

= 45 + 

tới đây tớ chịu , các cậu giúp với

Đầu tiên , cộng các phần nguyên lại với nhau , ta có :

( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\))

= 45 + \(\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)

sau khi cộng trong ngoặc , ta được 6 / 30 , rút gọn tối giản còn 1 / 5 

= 45 + \(\left(\frac{1}{5}+\frac{1}{20}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)

sau khi cộng trong ngoặc và rút gọn tối giản , ta được 1 / 4 

= 45 + \(\left(\frac{1}{4}+\frac{1}{2}\right)+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)

sau khi cộng trong ngoặc rồi rút gọn  , ta được 3 / 4

= 45 + \(\left(\frac{3}{4}+\frac{1}{12}\right)+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)

rút gọn lại ta được 5 / 6 

= 45 + \(\left(\frac{5}{6}+\frac{1}{42}\right)+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)

rút gọn tối giản ra 6 / 7

= 45 + \(\left(\frac{6}{7}+\frac{1}{56}\right)+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)

sau khi tính trong ngoặc rút gọn được 7 / 8

= 45 + \(\left(\frac{7}{8}+\frac{1}{72}\right)+\frac{1}{90}+\frac{1}{10}\)

tính trong ngoặc rồi rút gọn ra 8 / 9 

= 45 + \(\left(\frac{8}{9}+\frac{1}{90}\right)+\frac{1}{10}\)

cũng rút gọn tiếp ta được 9 / 10

= 45 + \(\left(\frac{9}{10}+\frac{1}{10}\right)\)

= 45 + 1

= 46

cái này thì bạn lên làm giảng viên đến nơi rồi

Dấu ^ là dấu gạch ngang của phản số nhé

a , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)  

  = \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)

b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)

 = \(\frac{1}{5}+0=\frac{1}{5}\)

c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)

 =  \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)

 = \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)

 = \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)

= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)

A 50

b 97

c 74

k cho mình nhé

Ta có: \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

= \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)

= \(\frac{9}{10}-\frac{1}{10}-\frac{1}{9}-...-\frac{1}{2}-\frac{1}{1}\)

= \(\frac{9}{10}+\frac{1}{10}-\frac{1}{1}\)

= 1 - 1 = 0

Vậy kết quả của phép tính là 0

Ta có : 
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2

= 9/10 -( 1/90 + 1/72 + ... + 1/2) 
= 9/10 - { 1/( 9.10) + 1/(9.8) + ... + 1/( 2.1)} 
= 9/10 - ( 1/9 - 1/10 + 1/8 - 1/9 + ...+ 1 - 1/2) ( 1/90 = 1/(9.10) = 1/9 - 1/10) 
= 9/10 - ( 1 - 1/10) 
= 9/10 - 9/10

= 0

9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2

=9/10-(1/9*10+1/8*9+...+1/1*2)

=9/10-(1/9-1/10+...+1-1/2)

=9/10-(-1/10+1)=9/10-9/10=0

= 9/1.10 + 1/9.10 + 1/8.9 + 1/7.8 + 1/6.7 +1/5.6 + 1/4.5 +1/3.4 +1/2.3 + 1/1.2

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 9/1.10 ( viết ngược lại)

= 1-1/2 + 1/2  -1/3 + 1/3 +....-1/10

= 1 - 1/10

= 9/10