Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=347\cdot4\cdot9\cdot400:8=347\cdot36\cdot50=624600\)
c: \(=16:\left\{400:\left[200-37-138\right]\right\}\)
\(=16:\left\{400:25\right\}=16:16=1\)
e: \(=46-\left[300:15\right]-2=46-20-2=24\)
\(\dfrac{1}{19}+\dfrac{9}{19\cdot29}+...+\dfrac{9}{1999\cdot2009}\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{1791}{38171}=\dfrac{200}{2009}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)}=\dfrac{-2}{3}\)
\(\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}-\dfrac{5}{22}\right)=\dfrac{-2}{3}}\)
\(=\dfrac{-3}{17}-\dfrac{2}{13}+\dfrac{20}{17}-\dfrac{11}{13}\)
=1-1
=0
\(-\left(\dfrac{3}{17}+\dfrac{2}{13}\right)-\left(\dfrac{-20}{17}+\dfrac{11}{13}\right)\)
\(=-\dfrac{3}{17}-\dfrac{2}{13}-\dfrac{20}{17}-\dfrac{11}{13}\)
\(=-\left(\dfrac{3}{17}+\dfrac{20}{17}\right)-\left(\dfrac{2}{13}+\dfrac{11}{13}\right)\)
\(=-\dfrac{23}{17}-1\)
\(=-\dfrac{40}{17}\)
\(a,\dfrac{3}{5}+\dfrac{1}{5}.\dfrac{-17}{9}=\dfrac{3}{5}-\dfrac{17}{45}=\dfrac{27}{45}-\dfrac{17}{45}=\dfrac{10}{45}=\dfrac{2}{9}\\ b,\left(-\dfrac{4}{15}-\dfrac{18}{19}\right)-\left(\dfrac{20}{19}+\dfrac{11}{15}\right)=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)-\left(\dfrac{18}{19}+\dfrac{20}{19}\right)=-1-2=-3\)
\(a,=\dfrac{3}{5}+\left(-\dfrac{17}{45}\right)=\dfrac{2}{9}\)
\(b,=-\dfrac{4}{15}-\dfrac{18}{19}-\dfrac{20}{19}-\dfrac{11}{15}=-3\)
\(0,125.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}\left(\dfrac{3}{7}-\dfrac{11}{7}\right)=\dfrac{1}{8}.-\dfrac{8}{7}=-\dfrac{1}{7}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Ta có:
\( \dfrac{7256.4375-725}{3650+4375.7255}\\ =\dfrac{7255.4375+4375-7255}{3650+4375.7255}\\ =\dfrac{7255.4375+\left(4375-725\right)}{3650+4375.7255}\\ =\dfrac{7255.4375+3650}{3650+4375.7255}\\ =1\)
Vậy giá trị của \(\dfrac{7256.4375-725}{3650+4375.7255}=1\)
7256.4375-725=(7255+1).4375-725=7255.4375+4375-725=7255.4375+3650
⇒(7256.4375−725)/(3650+4375.7255)=(7255.4375+3650)/(3650+4375.7255)=1