Bài 2: Tìm x,y a) \(-4,25-\left(x-\frac{5}{3}\right)_-< 3\frac{1}{2}-0,15\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+3^{10}.2^{20}}\)
\(=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
\(M=\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
\(=\frac{2^{19}\cdot\left(3^3\right)^3+3\cdot5\cdot\left(2^2\right)^9\cdot\left(3^2\right)^4}{6^9\cdot2^{10}+6^{10}\cdot2^{10}}\)
\(=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3\cdot3^8}{6^9\cdot2^{10}\left(6+1\right)}\)
\(=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3^9}{6^9\cdot2^{10}\cdot7}\)
\(=\frac{2^{18}\cdot3^9\left(2+5\right)}{2^{10}\cdot2^9\cdot3^9\cdot7}\)
\(=\frac{2^{18}\cdot3^9\cdot7}{2^{19}\cdot3^9\cdot7}\)
\(=\frac{1}{2}\)
\(\frac{9}{8}-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{72}=\frac{9}{8}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)
\(=\frac{9}{8}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
\(=\frac{9}{8}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{9}{8}-\left(1-\frac{1}{9}\right)=\frac{9}{8}-\frac{8}{9}=\frac{17}{72}\)
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{99.100}+\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)
\(=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)\)
\(=\frac{1}{100}-\left(-\frac{1}{100}+1\right)\)
\(=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=-\frac{49}{50}\)
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\frac{99}{100}\)
\(C=\frac{-98}{100}=\frac{-49}{50}\)
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
\(=\frac{1}{5}-\frac{1}{5}-\frac{3}{7}+\frac{3}{7}+\frac{5}{9}-\frac{5}{9}-\frac{2}{11}+\frac{2}{11}+\frac{7}{13}-\frac{7}{13}-\frac{9}{16}\)
\(=0+0+0+0+0-\frac{9}{16}\)
\(=\frac{-9}{16}\)