\(M=1\dfrac{1}{10}\cdot\dfrac{15}{19}-\dfrac{22}{38}\cdot0,5+110\%\cdot\dfrac{9}{19}\)

\(=\dfrac{11}{10}\cdot\dfrac{15}{19}-\dfrac{11}{19}\cdot\dfrac{1}{2}+\dfrac{11}{10}\cdot\dfrac{9}{19}\)

\(=\dfrac{15}{10}\cdot\dfrac{11}{19}-\dfrac{11}{19}\cdot\dfrac{1}{2}+\dfrac{9}{10}\cdot\dfrac{11}{19}\)

\(=\dfrac{11}{19}\left(\dfrac{15}{10}-\dfrac{1}{2}+\dfrac{9}{10}\right)\\ =\dfrac{11}{19}\cdot\dfrac{19}{10}\\ =\dfrac{11}{10}\)

\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)

Nhận xét

thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10

\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)

\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)

Mk bt giải bài này r nè! Thank ngonhuminh nhìu! Mk lm thêm câu Q, m.n xem thử mk lm cs đg k nha!!!!!

\(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...\left(\dfrac{19}{9}-1\right)\)

\(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...\left(\dfrac{9}{9}-1\right)...\left(\dfrac{19}{9}-1\right)\)

\(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...0...\left(\dfrac{19}{9}-1\right)\)

\(\Rightarrow Q=0\)

Bài 1:

\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)

\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)

\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)

a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)

\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)

\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)

\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\left(\dfrac{32}{60}+\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{37}{20}.\dfrac{24}{47}\\ =\dfrac{7}{5}+\dfrac{222}{235}\\ =\dfrac{329}{235}+\dfrac{222}{235}\\ =\dfrac{551}{235}\)

a: \(=\dfrac{2^{19}\cdot3^9+3^9\cdot5\cdot2^{18}}{2^{19}\cdot3^9+2^{10}}\)

\(=\dfrac{3^9\cdot2^{18}\cdot\left(2+5\right)}{2^{10}\cdot\left(2^9\cdot3^9+1\right)}=\dfrac{3^9\cdot7\cdot2^8}{6^9+1}\)

b: \(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}\cdot4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-29}{16}:\dfrac{-29}{16}=1\)

\(A=\dfrac{\left(1+17\right).\left(1+\dfrac{17}{2}\right)..........\left(1+\dfrac{17}{19}\right)}{\left(1+19\right).\left(1+\dfrac{19}{2}\right)..........\left(1+\dfrac{19}{17}\right)}\)

\(=\dfrac{18.\dfrac{19}{2}.............\dfrac{36}{19}}{20.\dfrac{21}{2}..........\dfrac{36}{17}}\)

\(=\dfrac{18.19.20.......36}{1.2.3...19}:\dfrac{20.21.....36}{1.2.3...17}\)

\(=\dfrac{1.2.3......36}{1.2.....36}\)

\(=1\)

\(\dfrac{5}{7} *\dfrac{-4}{19}-\dfrac{5*2}{7}*\dfrac{15}{19*2} =\dfrac{5}{7}*(\dfrac{-4}{19}-\dfrac{15}{19}) =\dfrac{5}{7}*(\dfrac{-19}{19})=\dfrac{-5}{7}\)

câu hỏi là gì

tối giản rồi mà bn!mẫu cx dương rồi!

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

\(=\dfrac{4}{3}\)

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

\(=-\dfrac{5}{11}\)