Tính giá trị của biểu thức :
A = \(\frac{1}{31}. [\frac{31}{5}.(9-\frac{1}{2})-\frac{17}{2}.(4+\frac{1}{5})]+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{930}\)
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Những câu hỏi liên quan
13 tháng 3 2019
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)
22 tháng 3 2020
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
15 tháng 3 2019
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
5 tháng 4 2017
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
14 tháng 3 2016
A = \(\frac{\frac{\frac{5+3.3-1.12}{12}}{3.6-5+2.2}+}{6}+\frac{16\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}=\frac{\frac{\frac{5+9-12}{12}}{18-5+4}}{6}+\frac{16}{17}=\frac{2}{12}.\frac{6}{17}+\frac{16}{17}=\frac{1}{17}.\frac{6}{17}=1\)
14 tháng 3 2016
A=\(\frac{\frac{5}{12}+\frac{9}{12}-\frac{12}{12}}{\frac{18}{6}-\frac{5}{6}+\frac{4}{6}}+\frac{16.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}\)
=\(\frac{\frac{1}{6}}{\frac{17}{6}}+\frac{16}{17}\)
=\(\frac{1}{17}+\frac{16}{17}\)
=1
M làm tiếp nha
A = 1/31 [ 3/15 ( 18/2 − 1/2 ) − 17/2 ( 20/5 + 1/5 ) + 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/30.31 ]
= 1/31 [3/15 . 17/2 − 17/2 . 21/5 + 1 − 1/2 + 1/2 − 1/3 + 1/3 − 1/4 + ... + 1/30 − 1/31 ]
= 1/31 [ 17/2 . ( 3/15−21/5 ) + 1 − 1/31]
= 1/31 [ 17/2 . 10/5 + 31/31 − 1/31 ]
= 1/31 [ 17/2 . 2 + 30/31 ]
= 1/31 [ 17 + 30/31 ]
= 1/31 [ 527/31 + 30/31 ]
= 1/31 . 557/31 = 557/961