= \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{90}\right)\)

= \(\left(1+1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)(9 số 1)

= 9 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)

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

= \(9+\left(1-\frac{1}{10}\right)=9+\frac{9}{10}=\frac{90}{10}+\frac{9}{10}=\frac{99}{10}\)

3/2+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=99/10=9,9

kết quả là 99/10 hoặc 9,9 nha bn

\(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)
=> \(2x+\frac{6+1}{6}+\frac{12+1}{12}+....+\frac{90+1}{90}=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+10=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=0\)
=>\(2x+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}=0\)
=>\(2x+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=0\)
=> \(2x-\frac{1}{10}=0\)
=>2x=\(\frac{1}{10}\)=> x=1/20
 

mình có bị nhầm chỗ dấu suy ra thứ 3. đáng lẽ ra biểu thức đó cộng 8 chứ k phải cộng 10 do mình sơ ý nên bạn hãy sủa lại chỗ ấy

đây nhá tại đánh máy lâu quá nên gửi hình ^^

\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)

\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)

\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)

\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}\)

\(\Rightarrow A=\frac{10}{11}\)

Ta có:

\(A=\frac{3}{2}+\frac{13}{12}+\frac{31}{30}+\frac{57}{56}+\frac{91}{90}\)

\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)

\(=\left(1+1+1+1+1\right)+\left(\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)\)

\(=5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\)

\(B=\frac{5}{6}+\frac{19}{20}+\frac{41}{42}+\frac{71}{72}+\frac{109}{110}\)

\(=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)

\(=\left(1+1+1+1+1\right)-\left(\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\right)\)

\(=5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\)

=> A - B =\(\left[5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\right]-\left[5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\right]\)

= \(5+\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}-5+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

= \(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)

\(B=\left(1-\frac{1}{6}\right)+\left(1-\frac{19}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)

Mk gợi ý đến đây thôi , mk bí rồi đợi mk nghĩ đã!

mk sửa lại 1-1/20 chứ ko phải 1-19/20

\(A=\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)\)

\(B=\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)

\(A-B=\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\right)\)

\(A-B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)

\(A-B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(A-B=1-\frac{1}{11}\)

\(A-B=\frac{10}{11}\)

Giải:

Vì

Nên ta phải chứng minh:

=> ( điều phải chứng minh)

Vì

Nên ta phải chứng minh:

=> ( điều phải chứng minh)

a) 

\(=\frac{7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10\cdot11\cdot11}{6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11\cdot10\cdot12}\)    

\(=\frac{7\cdot11}{6\cdot12}\)     

\(=\frac{77}{72}\)

b) 

\(=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)  

\(=6+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)  

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

\(=6+\frac{1}{2}-\frac{1}{8}\)  

\(=6+\frac{3}{8}\)

\(=\frac{51}{8}\)

Chia thành...a và b nhé.

Bg

a)Ta có: \(\frac{49}{48}.\frac{64}{63}.\frac{81}{80}.\frac{100}{99}.\frac{121}{120}\)

= \(\frac{49.64.81.100.121}{48.63.80.99.120}\)

= \(\frac{7.7.8.8.9.9.10.10.11.11}{6.8.7.9.8.10.9.11.10.12}\)

= \(\frac{7.11}{6.12}\)    (chịt tiêu trên dưới)

= \(\frac{77}{72}\)

b) Ta có: \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}\)

Có 6 số hạng  (đếm)

= \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)

= \(1+1+...+1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

= \(1.6+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

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

= \(6+\frac{1}{2}-\frac{1}{8}\)

= \(\frac{13}{2}-\frac{1}{8}\)

= \(\frac{51}{8}\)

Hơi dài....

THANK YOU