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

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

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

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

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

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

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

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

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

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

\(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}\)

\(=\frac{12-1}{12}+\frac{20-1}{20}+\frac{30-1}{30}+\frac{42-1}{42}+\frac{56-1}{56}+\frac{72-1}{72}+\frac{90-1}{90}+\frac{110-1}{110}\)

\(=1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}+1-\frac{1}{110}\)

\(=8-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

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

\(=8-\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{256}{33}\)

ĐẶT A = \(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{109}{110}\)

   8 -A = \(1-\frac{11}{12}+1-\frac{19}{20}+1-\frac{29}{30}+1-\frac{41}{42}+1-\frac{55}{56}+1-\frac{71}{72}+1-\frac{89}{90}+1-\frac{109}{110}\)   

8 -A   \(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\) 

8 -A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

8 - A = \(\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)

=> A = \(8-\frac{8}{33}=\frac{256}{33}\)

\(\frac{5^4.20^4}{25^5.4^5}\)

\(\frac{3}{2}\)+ \(\frac{7}{6}\)+\(\frac{13}{12}\)+\(\frac{21}{20}\)+\(\frac{31}{30}\)+ \(\frac{43}{42}\)+\(\frac{57}{56}\)+\(\frac{73}{72}\)+\(\frac{91}{90}\)

=\(\frac{3780+2940+2730+2646+2604+2580+2565+2555+2548}{2520}\)

= \(\frac{24948}{2520}\)=\(\frac{99}{10}\)

HỌC TỐT

Bằng 99/10

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

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

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

\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

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