A= 1/1.3.5+ 1/3.5.7+1/5.7.9+.....+1/95.97.99
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A = 1/4.( 4/1.3.5 + 4/3.5.7+ ....+ 4/95.97.99)
= 1/4 .( 1/ 1.3 - 1/3.5 + 1/3.5 - 1/5.7 + .......+ 1/95.97 - 1/97.99)
= 1/4( 1/1.3 - 1/97.99)
= 1/4 . 9499/29397
a) 9 + 99 + 999 + ... + 999999
= (10 - 1) + (100 - 1) + (1000 - 1) + ... + (1000000 - 1)
= (101 + 102 + 103 + ... + 106) - (1.6)
= 1111110 - 6 = 1111104
b) 1 + 11 + 111 + ... + 1111111
= 1 + (101 + 1) + (102 + 101 + 1) + ... + (106 + 105 + 104 + 103 + 102 + 101 + 1)
= 101 . 6 + 102 . 5 + 103 . 4 + ... + 106. 1) + (1 + 1.6)
= 60 + 500 + 4000 + ... + 1000000 + 7
= 1234560 + 7 = 1234567
c) C = 1.2 + 2.3 + 3.4 + 4.5 + ... + 98.99
3C = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 98.99.3
3C = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 98.99.(100 - 97)
3C = 1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 +...+ 98.99.100 - 98.99.97
3C = 98.99.100
C = \(\dfrac{98.99.100}{3}\) = 323400
d) D = 1.3.5 + 3.5.7 + 5.7.9 + ... + 95.97.99
8D = 1.3.5.8 + 3.5.7.8 + 5.7.9.8 + ... + 95.97.99.8
8D = 1.3.5.(7 + 1) + 3.5.7.(9 - 1) + 5.7.9.(11 - 3) + ... + 95.97.99.(101 - 93)
8D = 1.3.5.7 + 1.3.5.1 + 3.5.7.9 - 3.5.7.1 + 5.7.9.11 - 5.7.9.3 + ... + 95.97.99.101 - 95.97.99.93
8D = 1.3.5.1 + 95.97.99.101
D = \(\dfrac{1.3.5.1+95.97.99.101}{8}=15517600\)
1.3.5.8 + 3.5.7.8 + 5.7.9.8 + … + 95.97.99.8
= 1.3.5(7 + 1) + 3.5.7(9 - 1) + 5.7.9(11 - 3) + … + 95.97.99(101 - 93)
= 1.3.5.7 + 15 + 3.5.7.9 - 1.3.5.7 + 5.7.9.11 - 3.5.7.9 + … + 95.97.99.101 - 93.95.97.99
= 15 + 95.97.99.101
=> \(A=\frac{15.95+97.99.101}{8}\)
Đặt A = \(1.3.5+3.5.7+5.7.9+..+93.95.97+95.97.99\)
\(8A=1.3.5.8+3.5.7.8+...+93.95.97.8+95.97.99.8\)
\(8A=1.3.5.\left(1+7\right)+3.5.7.\left(9-1\right)+...+95.97.99\left(101-93\right)\)
\(8A=1.3.5.7+15+3.5.7.9-1.3.5.7+...+95.97.99.101-93.95.97.99\)
\(8A=15+95.97.99.101\)
\(\Rightarrow\) \(A=\frac{15+95.97.99.101}{8}\)
\(KL:........\)
\(G=1.3.5+3.5.7+5.7.9+...+95.97.99\)
\(G=1+99.\left(3+5+7+...+97\right)\)\
\(G=100.\left[\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\right]\)
\(G=100.\left(100.24\right)\)
\(G=100.2400=240000\)
a)\(A=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^2^5}\) <=>\(5A=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{24}}\)
<=>\(5A-A=(1+\frac{1}{5}+...+\frac{1}{5^{24}})-(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{25}})\)
<=>\(4A=1-\frac{1}{5^{25}}\) <=>\(A=\frac{(5^{25^{ }}-1)}{5^{25}}\div4\)
B = \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+....+\frac{1}{95.97.99}\)
B = \(\frac{1}{4}.\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+...+\frac{99-95}{95.97.99}\right)\)
B = \(\frac{1}{4}.\left(\frac{5}{1.3.5}-\frac{1}{1.3.5}+\frac{7}{3.5.7}-\frac{3}{3.5.7}+...+\frac{99}{95.97.99}-\frac{95}{95.97.99}\right)\)
B = \(\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
B = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{97.99}\right)\)
B = \(\frac{1}{4}.\frac{3200}{9603}\)
B = \(\frac{800}{9603}\)
Cho x và y thoả mãn (x-45)2=-|2y+5| tính giá trị của biểu thức: M=x2+y2+29/10.y-15
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=4.\frac{3200}{9603}=\frac{12800}{9603}\)
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\frac{3200}{9603}\)
\(A=\frac{800}{9603}\)
Bài trc mik làm lộn :)))
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