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Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
a) \(\frac{2}{3}=\frac{8}{12}\) ; \(\frac{1}{4}=\frac{3}{12}\)
mà 8 > 3 ⇒ \(\frac{8}{12}>\frac{3}{12}\)⇒\(\frac{2}{3}>\frac{1}{4}\)
b) \(\frac{7}{10}\) và \(\frac{7}{8}\); mà 10 > 8 ⇒ \(\frac{7}{10}< \frac{7}{8}\)
c) \(\frac{6}{7}=\frac{30}{35}\); \(\frac{3}{5}=\frac{21}{35}\)
mà 30 > 21 ⇒ \(\frac{30}{35}>\frac{21}{35}\)⇒\(\frac{6}{7}>\frac{3}{5}\)
d) \(\frac{14}{21}=\frac{2}{3}\); \(\frac{60}{72}=\frac{5}{6}\)
\(\frac{2}{3}=\frac{4}{6}\) ⇒ \(\frac{2}{3}< \frac{5}{6}\)⇒ \(\frac{14}{21}< \frac{60}{72}\)
e) \(\frac{38}{133}=\frac{2}{7}\); \(\frac{129}{344}=\frac{3}{8}\)
\(\frac{2}{7}=\frac{16}{56}\) ; \(\frac{3}{8}=\frac{21}{56}\) mà 16<21 ⇒ \(\frac{16}{56}< \frac{21}{56}\)⇒ \(\frac{38}{133}< \frac{129}{344}\)
f) \(\frac{11}{54}=\frac{22}{108}\)và \(\frac{22}{37}\) mà 108 > 37 ⇒ \(\frac{22}{108}< \frac{22}{37}\)⇒ \(\frac{11}{54}< \frac{22}{37}\)
Bài 1 :
a) \(A=\frac{-1}{4.5}+\frac{-1}{5.6}-\frac{-1}{7.8}+\frac{-1}{9.10}\)
\(A=\frac{1}{4}\)\(-\left(-\frac{1}{5}\right)+...+\left(-\frac{1}{9}\right)-\left(-\frac{1}{10}\right)\)
\(A=\frac{1}{4}+\frac{1}{10}\)
\(A=\frac{3}{20}\)
Bài 2:
a,17178585=1717:17178585:1717=15;13135151=1313:1015151:101=135115=51255<65255=1351⇒17178585<13135151a,17178585=1717:17178585:1717=15;13135151=1313:1015151:101=135115=51255<65255=1351⇒17178585<13135151
b,201201202202=201201:1001202202:1001=201202=201⋅1001001202⋅1001001=201201201202202202
Bài 2:
a, S = 1/11 + 1/12 + .. +1/20 với 1/2
SỐ số hạng tổng S: [20 - 11]: 1 + 1 = 10 số
mà 1/11 > 1/20
1/12 > 1/20
.........................
1/20 = 1/20
=> 1/11 + 1/12 + ... + 1/20 > 1/20 . 10 => S > 1/2
b, B = 2015/2016 + 2016/2017 và C = 2015+2016/2016+2017
Dễ dàng ta thấy: C = 4031/4033 < 1
B = 2015/2016 + 2016/2017
B = 2015/2016 + [1/2016 + 4062239/4066272]
B = [2015/2016 + 1/2016] + 4062239/4066272]
B = 1 +4062239/4066272
=> B > 1
Vậy B > C
c, [-1/5]^9 và [-1/25]^5
ta có: 255 = [52]5 = 52.5 = 510 > 59
=> [1/5]9 > [1/25]5
=> [-1/5]9 < [-1/25]5
d, 1/32+1/42+1/52+1/62 và 1/2
ta có: 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 = 1/9 + 1/16 + 1/25 + 1/36
mà: 1/9 < 1/8
1/16 < 1/8
1/25 < 1/8
1/36 < 1/8
=> 1/9+1/16+1/25+1/36 < 1/2
Vậy 1/32+1/42+1/52+1/62 < 1/2
Bài 1:
A = 3/4 . 8/9 . 15/16....2499/2500
A = [1.3/22][2.4/32]....[49.51/502]
A = [1.2.3.4.5...51 / 2.3.4....50][3.4.5...51 / 2.3.4...50]
A = 1/50 . 51/2
A = 51/100
B = 22/1.3 + 32/2.4 + ... + 502/49.51
B = 4/3.9/8....2500/2499
Nhận thấy B ngược A => B = 100/51 [cách tính tương tự tính A]
Bài 2:
a. S = 1/11+1/12+...+1/20 và 1/2
Số số hạng tổng S: [20 - 11]: 1 + 1 = 10 [ps]
ta có: 1/11 > 1/20