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TS
5
11 tháng 3 2023
A=
7.31
4
+
7.41
6
+
10.41
9
+
10.57
7
⇒
5
A
=
35.31
4
+
35.41
6
+
50.41
9
+
50.57
7
\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}
5
A
=
31
1
−
35
1
+
35
1
−
41
1
+
41
1
−
50
1
+
50
1
−
57
1
\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57}
5
A
=
31
1
−
57
1
=
31.57
26
⇒A=
31.57
130
PT
2
TB
15 tháng 6 2017
A= 5 x (4/31.7.5 + 6/7.5.41 + 9/41.10.5 + 7/ 10.5.57 + 1958/57.403.5)
= 5 x (4/31.35 + 6/35.41 + 9/41.50 + 7/50.57 + 1958/57.2015)
=5 x (1/31 - 1/35 + 1/35 - 1/41 + 1/41 - 1/50 + 1/50 - 1/57 + 1/57 - 1/2015)
= 5 x (1/31 - 1/2015)
= 5 x (65/2015 - 1/2015)
= 5 x 64/2015
= 320/2015 = 64/403
KK
0
MV
0
PH
1
24 tháng 5 2016
\(\frac{A}{5}=\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}=\frac{35-31}{35.31}+\frac{41-35}{35.41}+\frac{50-41}{50.41}+\frac{57-50}{50.57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)=> A = 5. \(\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(\frac{B}{2}=\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{38-31}{31.38}+\frac{43-38}{38.43}+\frac{46-43}{43.46}+\frac{57-46}{46.57}\)
\(\frac{B}{2}=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)=> B = 2.\(\left(\frac{1}{31}-\frac{1}{57}\right)\)
A/B = 5/2
NT
0
`Answer:`
\(A=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(=\frac{5.4}{5.\left(7.31\right)}+\frac{5.6}{5.\left(7.41\right)}+\frac{5.9}{5.\left(10.41\right)}+\frac{5.17}{5.\left(10.57\right)}\)
\(=\frac{20}{35.31}+\frac{30}{35.41}+\frac{45}{50.41}+\frac{35}{50.57}\)
\(=5.\left(\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}\right)\)
\(=5.\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)
\(=5.\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(=5.\frac{26}{1767}\)
\(=\frac{130}{1767}\)