Cho A = 1/72 - 1/74 + 1/76 - 1/78 +.....+ 1/798 - 1/7100
CMR : A < 1/50
Giúp mình vs, mk đang cần . thank you các bạn
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Bài 1: 100 - 94 + 90 - 84 + 80 - 74 + 70 - 64 + 60 - 50 =34
Bài 2: 78 . 31 + 78 . 24 + 78 . 17 + 22 . 72= 7200
Bài 3: 2 . 450 . 25 . 8 =180000
Bài 1: 100 - 94 + 90 - 84 + 80 - 74 + 70 - 64 + 60 - 50
=6+6+6+6+6=6x5=30
Bài 2: 78 . 31 + 78 . 24 + 78 . 17 + 22 . 72
=78x(31+24+17+22)
=...
Bài 3: 2 . 450 . 25 . 8 =450x200=900000
A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\)...\(\dfrac{9999}{10000}\)
A = \(\dfrac{1.3.2.4..3.5......99.101}{2.2.3.3.4.4....100.100}\)
A = \(\dfrac{1.2.3..4.5.....99}{2.3.4.5.....99.100}\).\(\dfrac{3.4.5....100.101}{2.3.4.5...100}\)
A = \(\dfrac{1}{100}\).\(\dfrac{101}{2}\)
A = \(\dfrac{101}{200}\)
2; B = (1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{8}\))...(1 - \(\dfrac{1}{n+1}\))
Xem lại đề bài.
A)10+32+54+76+98=270
B)54+90+36+12+78=270
C)74+18+92+30+56=270
\(\Rightarrow\)A=B=C
Co: A=10 + 32 + 76 + 98
A= (10 + 98) + (32 + 76)
A= 108 + 108
A= 108 . 2
A= 216
Co: B= 54 + 90 + 36 + 12 + 78
B= (54 + 36) + (12 + 78) + 90
B= 90 + 90 + 90
B= 90 . 3
B= 270
Co: C= 74 + 18 + 92 + 30 + 56
C= (74 + 18) + (30 + 56) + 92
C= 92 + 92 + 86
C= 92 . 2 + 86
C= 184 + 86
C= 270
=> 216 < 270 = 270
Vay: A > B = C
\(\left(1+\frac{1}{11}\right)\left(1+\frac{1}{12}\right)...\left(1+\frac{1}{16}\right)\)
\(=\frac{12.13.14....17}{11.12.13....16}=\frac{17}{11}\)
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)\(+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)\(+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
=1-1/9
=8/9
a) \(-\left(76-35+15\right)+\left[18-27+\left(-17\right)\right]\)
\(=-76+35+15+18-27-17\)
\(=-52\)
b) \(167+\left[127-235-\left(-16\right)\right]+\left(-67\right)\)
\(=167+127-235+16-67\)
\(=8\)
c) \(\left(-19\right)+165-\left[27+\left(-21\right)-\left(+72\right)\right]\)
\(=\left(-19\right)+165-27+21+72\)
\(=212\)
d) \(89.\left(-2\right)+\left[20+\left(-2\right).\left(-5\right)-85\right]\)
\(=\left(-178\right)+\left(20+10-85\right)\)
\(=\left(-178\right)+20+10-85\)
\(=-233\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)
Vậy \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(\Rightarrow A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)\(\left(dpcm\right)\)
Bài này lớp 6 mik kiểm tra chất lượng đầu năm nè
\(\frac{1}{7^2}A=\frac{1}{7^2}\left(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\right)\)
\(\Leftrightarrow\frac{1}{7^2}A=\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-\frac{1}{7^{10}}+...+\frac{1}{7^{100}}-\frac{1}{7^{102}}\)
\(\Leftrightarrow A+\frac{1}{7^2}A=\frac{1}{49}-\frac{1}{7^{102}}\Rightarrow\frac{50}{49}A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow A=\left(\frac{1}{49}-\frac{1}{7^{102}}\right)\cdot\frac{49}{50}< \frac{1}{50}\left(đpcm\right)\)