Tính giá trị biểu thức:
A=1/31[31/5(9−1/2)−17/2(4+1/5)+1/2+1/6+1/12+...+1/930]\
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T
0
25 tháng 2 2022
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
13 tháng 3 2019
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)
3 tháng 7 2018
\(-\frac{1}{10}< =x< =\frac{3}{5}\)
\(\frac{-4}{9}< x< =\frac{2}{3}\)
22 tháng 3 2020
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 7 2024
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 7 2024
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
15 tháng 8 2018
(1) 5/2 x (3/4 - 1/3) -11/12+1/4 = 5/2 x (9/12-4/12) -11/12 +3/12=5/2 x 5/12 -11/12 +3/12 = 25/24 - 11/12 +3/12=25/24 -22/24 + 6/24=9/12=3/4
(2) 11/12 : (4/10 + 3/5) + (5/6 - 1/2) x2/3
= 11/12 : (4/10 + 6/10) + (5/6-3/6) x 2/3
= 11/12:1 +1/2 x 2/3
= 11/12 + 2/3 =11/12+8/12 = 19/12
Kết quả hơi lớn bạn nhé!
A=\(\frac{1}{31}\left[\frac{31}{5}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{930}\right]\)
=\(\frac{1}{31}\left[\frac{31}{5}\left(\frac{18}{2}-\frac{1}{2}\right)-\frac{17}{2}\left(\frac{20}{5}+\frac{1}{5}\right)+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{30.31}\right]\)
=\(\frac{1}{31}\left[\frac{31}{5}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{30}-\frac{1}{31}\right]\)
=\(\frac{1}{31}\left[\frac{17}{2}.\left(\frac{31}{5}-\frac{21}{5}\right)+1-\frac{1}{31}\right]\)
=\(\frac{1}{31}\left[\frac{17}{2}.\frac{10}{5}+\frac{31}{31}-\frac{1}{31}\right]\)
=\(\frac{1}{31}\left[\frac{17}{2}.2+\frac{30}{31}\right]\)
=\(\frac{1}{31}\left[17+\frac{30}{31}\right]\)
=\(\frac{1}{31}\left[\frac{527}{31}+\frac{30}{31}\right]\)
=\(\frac{1}{31}.\frac{557}{31}=\frac{557}{961}\)
A = bao nhiêu vậy bạn