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Hôm nay olm.vn sẽ hướng dẫn các em giải dạng toán nâng cao cấu trúc thi hsg, violympic  tính nhanh tổng các phân số mà tử số bằng tổng hai thừa số dưới mẫu:

Kiến thức cần nhớ:  \(\dfrac{a}{b.c}\)  trong đó a = b + c

+ Tách từng phân số thanh tổng hai phân số :

 \(\dfrac{a}{b.c}\) = \(\dfrac{b+c}{b.c}\) = \(\dfrac{1}{b}+\dfrac{1}{c}\)

+ Triệt tiêu các hạng tử giống nhau.

+ Thu gọn ta được tổng cần tính                               

  \(\dfrac{2}{6}+\dfrac{9}{20}-\dfrac{11}{30}+\dfrac{13}{42}-\dfrac{15}{56}+\dfrac{17}{72}-\dfrac{19}{90}+\dfrac{21}{110}-\dfrac{23}{132}\)

=\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)-(\(\dfrac{1}{5}+\dfrac{1}{6}\))+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)-(\(\dfrac{1}{7}\)+\(\dfrac{1}{8}\))+\(\dfrac{1}{8}\) +\(\dfrac{1}{9}\)-(\(\dfrac{1}{9}\)+\(\dfrac{1}{10}\))+\(\dfrac{1}{10}\)+\(\dfrac{1}{11}\)-(\(\dfrac{1}{11}\)+\(\dfrac{1}{12}\))

= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)

= \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)

= \(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{1}{12}\)

= \(\dfrac{1}{2}\)

a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)

Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)

b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)

\(\dfrac{25}{49}=1-\dfrac{24}{49}\)

Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)

Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)

23/27 lớn hơn 22/29

15/25 lớn hơn 25/49

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

\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)

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

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)

b) Bạn làm tương tự. 

mnbvcxzasdf

976554

Bạn làm phân số bù đi bài này dễ lắm!!!!

nhưng con c bài 1 làm phần bù thế nào hả bạn

cảm ơn bạn

Tìm x biết:

\(\frac{x}{3}-\frac{3}{4}=\frac{1}{12}\)

\(\frac{x}{3}=\frac{1}{12}+\frac{3}{4}\)

\(\frac{x}{3}=\frac{5}{6}\)

\(x=\frac{5}{6}.3\)

\(x=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}\)

\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)

\(\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)

\(\frac{13}{23}+x=\frac{199}{230}\)

\(x=\frac{199}{230}-\frac{13}{23}\)

\(x=\frac{3}{10}\)

Vậy \(x=\frac{3}{10}\)

Bài 2: tính

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

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

\(=\frac{1}{5}-\frac{1}{11}\)

\(=\frac{6}{55}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{1}-\frac{1}{50}\)

\(=\frac{49}{50}\)

Bài 2:

1/30+1/42+1/56+1/72+1/90+1/110

=1/5.6+1/6.7+1/7.8+1/8.9+1/9.10+1/10.11

=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11

=1/5-1/11=6/55

b)1/1.2+1/2.3+...+1/49.50

=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50

=49/50