a) \(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{120}{121}.\dfrac{143}{144}\)

= \(\dfrac{1.3.2.4.3.5.4.6....10.12.11.13}{2^2.3^2.4^2.5^2...11^2.12^2}\)

= \(\dfrac{1.2.12.13}{2^2.12^2}=\dfrac{13}{2.12}=\dfrac{13}{24}\)

b) \(B=\dfrac{5}{9}.\dfrac{21}{25}.\dfrac{45}{49}.\dfrac{77}{81}....\dfrac{357}{361}.\dfrac{437}{441}\)

= \(\dfrac{1.5.3.7.5.9.7.11.....17.21.19.23}{3^2.5^2.7^2....19^2.21^2}=\dfrac{1.3.21.23}{3^2.21^2}\)

= \(\dfrac{23}{3.21}=\dfrac{23}{63}\)

Ta có: \(S=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)

\(=\dfrac{3}{2^2}\cdot\dfrac{2^3}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{99}{10^2}\)

\(=\dfrac{11}{20}\)

b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)

Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé

9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)

10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)

\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)

\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)

\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)

\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)

\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)

\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)

\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)

\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)

Giải:

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)

\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)

\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)

\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)

Vậy \(A=\dfrac{30}{31}.\)

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}............................\dfrac{899}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}..............................\dfrac{29.31}{30^2}\)

\(\Leftrightarrow A=\dfrac{1.3.2.4.3.5..........29.31}{2.2.3.3.4.4.........30.30}\)

\(\Leftrightarrow A=\dfrac{\left(2.3.........29.30\right).\left(3.4.5......29.31\right)}{\left(2.3....29.30\right).\left(2.3.4.......29.30\right)}\)

\(\Leftrightarrow A=\dfrac{31}{2.30}=\dfrac{31}{60}\)

\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}....................................\dfrac{2499}{2500}\)

\(\Leftrightarrow B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.............................\dfrac{49.51}{50.50}\)

\(\Leftrightarrow B=\dfrac{\left(2.3.4.....49\right).\left(4.5.6......51\right)}{\left(3.4.5....50\right)\left(3.4.5.....50\right)}=\dfrac{2.51}{50.3}=\dfrac{17}{25}\)

a,\(3\dfrac{17}{24}+\left(2\dfrac{8}{15}-4\dfrac{8}{15}\right):\left(2\dfrac{11}{30}-\dfrac{11}{30}\right)\)

\(=\dfrac{89}{24}-2:2\)

\(=\dfrac{65}{24}\)

b,\(0,5:\sqrt{625}-\sqrt{\dfrac{4}{25}}+0,18.\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right)\)

\(=0,5:25-\dfrac{2}{5}+0,18.\dfrac{1}{2}\)

\(=-\dfrac{29}{100}\)