Tính:
a,A= 3.4+4.5+5.6+...+49.50
b,B=1.3+3.5+5.7+...+51.53
c,C=5.5+6.6+...+30.30
d,D=1+3+9+27+81+243+729+218+6561
Dấu"." là dấu nhân các bạn nhé giúp mk gấp nhé ai nhanh mk k 5 cái
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10 tháng 7 2018
a) A = 3.4 + 4.5 + 5.6 + ...+ 49.50
=> 3A = 3.4.3+4.5.3+ 5.6.3+...+49.60.3
3A = 3.4.(5-2) +4.5.(6-3) + 5.6.(7-4) + ...+ 49.60.(61-48)
3A = 3.4.5 - 2.3.4 + 4.5.6 -3.4.5 + 5.6.7-4.5.6 + 49.60.61 - 48.49.60
3A = -2.3.4 + 49.60.61
\(A=\frac{-2.3.4+49.60.61}{3}=59772\)
b) B = 1.3 + 3.5 + 5.7 + ...+ 51.53
=> 6B = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 51.53.6
6B = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) +...+ 51.53.(55-49)
6B = 1.3.5 + 1.3 + 3.5.6 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 51.53.55 - 49.51.53
6B = 1.3 + 51.53.55
\(B=\frac{1.3+51.53.55}{6}=24778\)
10 tháng 7 2018
cau c mk ko bk
d) D = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
D = 30+31+32+33+34+35+36+37+38
=> 3D = 31+32+33+...+38+39
=> 3D - D = 39-30
2D = 39-1
\(D=\frac{3^9-1}{2}=9841\)
CA
7 tháng 6 2018
a,0,36.350+1,2.20.3+9.4.4,5
=13.3.35+12.2.3+9.2.3.3
=3.(13.35+12.2+.9.2.3)
=3.(455+24+54)
=3.533
=1599
b,2015.2016-5/2015.2015+2010
=4062240-5+2010
=4064245
c,2/1.3+2/3.5+2/5.7+...+2/71.73
=1-1/3+1/3-1/5+1/5-1/7+...+1/71-1/73
=1-1/73
=72/73
d,(1+1/2).(1+1/3)+...+(1+1/2018)
=3/2.4/3.5/4+...+2019/2018
=2019/2
e,E=1/4.5+1/5.6+1/6.7+...+1/80.81(làm tương tự với phần d nên mình làm ngắn
=1/4-1/81
=77/324
f,F=3/2.3+3/3.4+...+3/99.100
=3.(1/2.3+1/3.4+...+1/99.100)(làm tương tự với d
=3.(1/2-1/100)
=3.49/100
=147/100
gG=5/1.4+5/4.7+...+5/61.64
3G=5.(3/1.4+3./4.7+...+3/61.64)
=5.(1-1/64)
=5.63/64
=315/64
ok nha bạn,mình giữ đúng lời hứa.
31 tháng 1 2021
Bài 5:
a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)
\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)
\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)
hay A=330
Vậy: A=330
AH
15 tháng 8 2016
- \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
IM
20 tháng 8 2016
b)
\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)
\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)
\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)
\(=\frac{406}{183}\)
d)
\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)
\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)
=57/37
PH
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 6 2023
(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))
(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))
Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)
Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2
Dãy số trên có số số hạng là: (25 - 1): 2 + 1 = 13
Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)
A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))
A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)( \(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))
A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)
A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)
Đặt B = \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\)
B\(\times\)3 =1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)
B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)
2B = \(\dfrac{242}{243}\)
B = \(\dfrac{242}{243}\): 2
B = \(\dfrac{121}{243}\)
11a + B = 11a + \(\dfrac{121}{243}\) (2)
Từ (1) và(2) ta có:
a\(\times\)13 + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)
a \(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\)
\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)
a \(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)
a \(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)
a \(\times\) 2 = \(\dfrac{109}{6075}\)
a = \(\dfrac{109}{6075}\): 2
a = \(\dfrac{109}{12150}\)
