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\(A=\frac{2.3.5+4.9.25+6.9.35+10.21.40}{2.3.7+4.9.35+6.9.49+10.21.56}\)
\(A=\frac{\left(2.3.5\right)+\left(2.3.5\right).2.3.5+\left(2.3.5\right).3.3.7+\left(2.3.5\right).5.7.8}{\left(2.3.7\right)+\left(2.3.7\right).2.3.5+\left(2.3.7\right).3.3.7+\left(2.3.7\right).5.7.8}\)
\(A=\frac{\left(2.3.5\right).\left(1+2.3.5+3.3.7+5.7.8\right)}{\left(2.3.7\right).\left(1+2.3.5+3.3.7+5.7.8\right)}\)
\(A=\frac{2.3.5}{2.3.7}=\frac{5}{7}.\)
\(B=\left(-\frac{3}{4}\right).\left(-\frac{8}{9}\right).\left(-\frac{15}{16}\right)...\left(-\frac{399}{400}\right)\)
\(B=-\frac{1.3.2.4.3.5...19.21}{2.2.3.3.4.4...20.20}\)
\(B=-\frac{1.2.3...19.3.4.5...21}{2.3.4...20.2.3.4...20}=-\frac{21}{40}.\)
\(A=\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot...\cdot\frac{360}{361}\cdot\frac{399}{400}\)
\(A=\frac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot18\cdot20\cdot19\cdot21}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot19\cdot19\cdot20\cdot20}\)
\(A=\frac{2\cdot21}{3\cdot20}\)
\(A=\frac{7}{10}\)
\(B=\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{441}{440}\cdot\frac{484}{483}\)
\(B=\frac{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot21\cdot21\cdot22\cdot22}{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot20\cdot22\cdot21\cdot23}\)
\(B=\frac{3\cdot22}{2\cdot23}=\frac{33}{23}\)
\(C=\frac{17}{23}.\left(\frac{7}{61}+\frac{28}{61}+\frac{26}{61}\right)\)
\(C=\frac{17}{23}\cdot1=\frac{17}{23}\)
1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
Câu 17:
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2006}\right).\left(1+\frac{1}{2007}\right)\)
=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2007}{2006}.\frac{2008}{2007}\)
\(=\frac{2008}{2}=1004\)
Câu 18:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2006}\right).\left(1-\frac{1}{2007}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2005}{2006}.\frac{2006}{2007}\)
\(=\frac{1}{2007}\)
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
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{399}{400}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{19.21}{20.20}\)
\(=\frac{1.2.3.....19}{2.3.4.....20}.\frac{3.4.5.....21}{2.3.4.....20}\)
\(=\frac{1}{20}.\frac{21}{2}\)
\(=\frac{21}{40}\)
\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)
= \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{399}{400}\)
= \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{19.21}{20.20}\)
= \(\frac{1.2.3...19}{2.3.4...20}.\frac{3.4.5...21}{2.3.4...20}\)
= \(\frac{1}{20}.\frac{21}{2}=\frac{21}{40}\)