\(\left[\frac{31}{\left(2\frac{1}{3}\right):4}:\frac{31}{11}\right]\frac{1}{12}.x=\frac{1}{16}\)
minh can cach lam
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Những câu hỏi liên quan
Tinh bang cach hop li
a) \(\left(\frac{1}{4}+\frac{-5}{13}\right)\)+ \(\left(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4}\right)\)
b)\(\left(\frac{21}{31}+\frac{-16}{7}\right)\)+\(\left(\frac{44}{53}+\frac{10}{31}\right)+\frac{9}{53}\)
a)(1/4+3/4)-(5/13+8/13)+2/11=1-1+2/11=2/11
b)(21/31+10/31)-16/7+(44/53+9/53)=1+1-16/7=14/7-16/7=-2/7
bằng2/11
-16/7
a, \(\left(\frac{1}{4}+\frac{-5}{13}\right)+\left(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4}\right)\)
\(=\left(\frac{1}{4}-\frac{5}{13}\right)+\left(\frac{2}{11}-\frac{8}{13}+\frac{3}{4}\right)\)
\(=\frac{1}{4}-\frac{5}{13}+\frac{2}{11}-\frac{8}{13}+\frac{3}{4}=1-1+\frac{2}{11}=\frac{2}{11}\)( vì \(\left(\frac{1}{4}+\frac{3}{4}\right)+\left(-\frac{5}{13}-\frac{8}{13}+\frac{2}{11}\right)\))
b, \(\left(\frac{21}{31}+\frac{-16}{7}\right)+\left(\frac{44}{53}+\frac{10}{31}\right)+\frac{9}{53}\)
\(=\frac{21}{31}-\frac{16}{7}+\frac{44}{53}+\frac{10}{31}+\frac{9}{53}\)
\(=1-\frac{16}{7}+1=\frac{7}{7}-\frac{16}{7}+\frac{7}{7}=-\frac{2}{7}\)
\([\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)].\frac{31}{50}\)
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
Tính \(E=\left[\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)\right].\frac{31}{50}\)
Tính
\(\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15.6\frac{1}{3}.\frac{2}{19}\right)}{-4\frac{5}{6}+\frac{1}{6}.\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right).\frac{31}{56}\)
1)\(\left[\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)\right].\frac{31}{50}\)
Tính giá trị biểu thức:
\(A=\left[\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)\right].\frac{31}{50}\)
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
\(a.A=[\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}]+\frac{1890}{2005}+115\)
b.B=\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\cdot\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(42-5\frac{1}{3}\right)}\cdot\left(-1\frac{19}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{1\frac{11}{31}.4\frac{3}{7}-\left(15-6\frac{1}{3}.\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}.\left(-1\frac{14}{93}\right)\)tất cả .\(\frac{31}{50}\)
Bài 4 :
a) Tính giá trị của biểu thức :
\(A=\left(\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right)\cdot\frac{31}{50}\)
b) Chứng tỏ rằng : \(B=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{3^2}-...-\frac{1}{2004^2}>\frac{1}{2004}\)