Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{-19}{9}.\dfrac{1}{2}-\dfrac{4}{11}.\dfrac{-11}{9}+\left(-\dfrac{2}{3}\right)=-\dfrac{23}{18}\)
\(B=\left(-\dfrac{15}{6}\right):\dfrac{-1}{2}+\dfrac{7}{-12}-\dfrac{1}{3}.\dfrac{-11}{2}=\dfrac{25}{4}\)
\(C=\dfrac{3}{4}.\left(-8\right)-\dfrac{1}{3}.\dfrac{-7}{2}-\dfrac{5}{18}=-\dfrac{46}{9}\)
\(A=\dfrac{-19}{18}+\dfrac{4}{9}-\dfrac{2}{3}=\dfrac{-19}{18}+\dfrac{8}{18}-\dfrac{12}{18}=\dfrac{-23}{18}\)
\(B=\dfrac{-5}{2}\cdot\dfrac{-2}{1}-\dfrac{7}{12}+\dfrac{11}{6}=\dfrac{5\cdot12-7+22}{12}=\dfrac{75}{12}=\dfrac{25}{4}\)
\(\left(1+\dfrac{7}{9}\right).\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}.\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
\(=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.\dfrac{55}{48}...\dfrac{7}{180}\)
\(=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.\dfrac{5.11}{4.12}...\dfrac{11.17}{10.18}\)
\(=\dfrac{\left(2.3.4.5...11\right).\left(8.9.10.11...17\right)}{\left(1.2.3.4...10\right).\left(9.10.11.12...18\right)}\)
\(=\dfrac{11.8}{1.18}=\dfrac{88}{18}=\dfrac{44}{9}\)
ta có ;
\(\left(1+\dfrac{7}{9}\right)\cdot\left(1+\dfrac{7}{20}\right).\left(1+\dfrac{7}{33}\right)...\left(1+\dfrac{1}{180}\right)\)
=\(\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}....\dfrac{187}{180}\)
=\(\dfrac{8.2}{9.1}.\dfrac{9.3}{10.2}.\dfrac{10.4}{3.11}.\dfrac{11.5}{4.12}....\dfrac{17.11}{18.10}\)
=\(\dfrac{8.9.10.11.12.13.14.15.16.17.2.3.4.5.6.7.8.9.10.11}{9.10.11.12.13.14.15.16.17.18.1.2.3.4.5.6.7.8.9.10}\)
=\(\dfrac{8.11}{18}=\dfrac{88}{18}=\dfrac{44}{9}\)
\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Vậy T = 50
\(T=\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{98}+1\right)\cdot\left(\dfrac{1}{99}+1\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}+\dfrac{4}{4}\right)\cdot...\cdot\left(\dfrac{1}{98}+\dfrac{98}{98}\right)\cdot\left(\dfrac{1}{99}+\dfrac{99}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{2\cdot3\cdot4\cdot...\cdot98\cdot99}\)
\(=\dfrac{100}{2}=50\).
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
a)\(\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right)\cdot\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}\cdot\dfrac{-5}{6}+\left(\dfrac{-1}{4}\right)=\dfrac{5}{12}+\dfrac{1}{2}=\dfrac{11}{12}\)
b)\(17\dfrac{11}{9}-\left(6\dfrac{3}{13}+7\dfrac{11}{19}\right)+\left(10\dfrac{3}{13}-5\dfrac{1}{4}\right)=\dfrac{164}{9}-\left(\dfrac{81}{13}+\dfrac{144}{19}\right)+\left(\dfrac{133}{13}-\dfrac{21}{4}\right)=\dfrac{164}{9}-\dfrac{3411}{247}+\dfrac{259}{52}=\dfrac{6425}{684}\)
c)\(\left(\dfrac{-3}{2}\right)^2-\left[-2\dfrac{1}{3}-\left(\dfrac{3}{4}+\dfrac{1}{3}\right):2\dfrac{3}{5}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left[\dfrac{-7}{3}-\dfrac{13}{12}\cdot\dfrac{5}{13}\right]\cdot\left(\dfrac{-3}{4}\right)=\dfrac{9}{4}-\left(\dfrac{-11}{4}\right)\cdot\left(\dfrac{-3}{4}\right)=\dfrac{3}{16}\)
d)\(\dfrac{21}{33}:\dfrac{11}{5}-\dfrac{13}{33}:\dfrac{11}{5}+\dfrac{25}{33}:\dfrac{11}{5}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{21}{33}-\dfrac{13}{33}+\dfrac{25}{33}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=1\)
\(a)\dfrac{2}{3}-\dfrac{3}{5}:\left(-1\dfrac{1}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{3}{5}:\left(\dfrac{-6}{5}\right)+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\left(\dfrac{-2}{3}\right).\dfrac{3}{8}\)
\(=\dfrac{2}{3}-\dfrac{-1}{2}+\dfrac{-1}{4}\)
\(=\dfrac{7}{6}+\dfrac{-1}{4}\)
\(=\dfrac{11}{12}\)
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
\(T=\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).......\left(\dfrac{1}{98}+1\right).\left(\dfrac{1}{99}+1\right) \) \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}....\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
\(T=\left|\dfrac{1}{2}+1\right|\left|\dfrac{1}{3}+1\right|\left|\dfrac{1}{4}+1\right|.....\left|\dfrac{1}{98}+1\right|\left|\dfrac{1}{99}+1\right|\)
\(T=\left|\dfrac{3}{2}\right|.\left|\dfrac{4}{3}\right|.\left|\dfrac{5}{4}\right|......\left|\dfrac{99}{98}\right|.\left|\dfrac{100}{99}\right|\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5.....99.100}{2.3.4.....98.99}=\dfrac{100}{2}=50\)