\(=\left[6\cdot\dfrac{1}{9}+1+1\right]\cdot\left(-3\right)-1\)
=(2/3+2)x(-3)-1
\(=\dfrac{8}{3}\cdot\left(-3\right)-1=-8-1=-9\)
Đúng 0
Bình luận (0)
Violympic toán 7
Các câu hỏi tương tự
\(tính:\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2}\)
2 tháng 7 2021 lúc 21:10
Tính giá trị biểu thức A , biết rằng A M : N Mà M dfrac{dfrac{1}{99}+dfrac{2}{98}+dfrac{3}{97}+dfrac{4}{96}+...+dfrac{97}{3}+dfrac{98}{2}+dfrac{99}{1}}{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{100}} N dfrac{92-dfrac{1}{9}-dfrac{2}{10}-dfrac{3}{11}-...-dfrac{90}{98}-dfrac{91}{99}-dfrac{92}{100}}{dfrac{1}{45}+dfrac{1}{50}+dfrac{1}{55}+...+dfrac{1}{495}+dfrac{1}{500}}
Đọc tiếp
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(\left[18\dfrac{1}{6}-\left(0,06:7\dfrac{1}{2}+3\dfrac{2}{5}\cdot0,38\right)\right]:\left(19-2\dfrac{2}{3}\cdot4\dfrac{3}{4}\right).Tính\)
8) Adfrac{9}{10}-dfrac{1}{90}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-dfrac{1}{30}-dfrac{1}{20}-dfrac{1}{12}-dfrac{1}{6}-dfrac{1}{2}
9) Bdfrac{1}{3}+dfrac{1}{3^2}+dfrac{1}{3^4}+...+dfrac{1}{3^{2014}}+dfrac{1}{3^{2015}}
10) Pdfrac{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2005}}{dfrac{2004}{1}+dfrac{2003}{2}+dfrac{2002}{3}+...+dfrac{1}{2004}}
Đọc tiếp
8) \(A=\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
9) \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2014}}+\dfrac{1}{3^{2015}}\)
10) \(P=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2005}}{\dfrac{2004}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+...+\dfrac{1}{2004}}\)
a,\(-4\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{6}\right)< hoac=x< hoac=\dfrac{-2}{5}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
b, \(-4\dfrac{2}{5}.2\dfrac{4}{3}< hoac=x< hoac=-2\dfrac{3}{5}:1\dfrac{6}{15}\)
Câu 1: Tính
1) \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
2) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{1}{2}-0,25+0,2}{\dfrac{7}{6}-0,875+0,7}+\dfrac{6}{7}\)
Tìm số nguyên x biết: a) \(-4\dfrac{3}{5}.2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
b)\(-4\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\le x\le-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
tính: A=\([\dfrac{1\dfrac{11}{31}.4\dfrac{3}{7}-(15-6\dfrac{1}{3}.\dfrac{2}{19})}{4\dfrac{5}{6}+\dfrac{1}{6}(12-5\dfrac{1}{3})}.(-1\dfrac{14}{93})].\dfrac{31}{50}\)
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{10}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(\dfrac{\left(1+2+3+...+99+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}\)
\(\dfrac{(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}).230.\dfrac{1}{25}+46\dfrac{3}{4}}{(1\dfrac{3}{10}+\dfrac{10}{3}):(12\dfrac{1}{3}-14\dfrac{2}{7})}\)