Tính:A: 1+1/3+1/6+1/10+...+1/171+1/190.
B: S=1/2+1/2^2+1/2^3+...+1/2^20.
C: 1+2+2^2+2^3+...+2^2006+2^2007.
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\(A=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{380}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{19.20}\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=1+2\times\frac{9}{20}\)
\(=1+\frac{9}{10}\)
\(=\frac{19}{10}\)
b)\(2S=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)
\(2S=1+\frac{1}{2}+...+\frac{1}{2^{19}}\)
\(2S-S=\left(1+\frac{1}{2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)
\(S=1-\frac{1}{2^{20}}\)
c)đặt A=1+2+2^2+2^3+...+2^2006+2^2007.
2A=2(1+2+2^2+2^3+...+2^2006+2^2007)
2A=2+2^2+2^3+...+2^2008
2A-A=(2+2^2+2^3+...+2^2008)-(1+2+2^2+2^3+...+2^2006+2^2007)
A=2^2008-1
1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)
=(1-1/3)....0.....(1-9/5)
=0
=>đpcm.
b)ta xét:
1/22 = 1/2x2 < 1/1x2
.............
1/82 = 1/8x8 <1/7x8
=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8
<=> B <1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8
<=> B < 1 - 1/8 = 7/8 < 1
=> B < 1 => đpcm
2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)
Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)
Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)
=> A > B
b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C
=> C > D
c)gọi 2010 là a
ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)
áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)
=> E > F
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot......\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot......\cdot\frac{19}{20}\)
\(A=\frac{1.2.3.....19}{2.3........20}\)
\(A=\frac{1}{20}\)
1. 2006/987654321 + 2007/246813579 = 2007/246813579 + 2006/987654321
=>
2.
3 - (5.3/8 + X - 7 . 5/24) : 6 . 2/3 =2
3 - (15/8 + X - 35/24) : 4 = 2
3 - (15/8 + X - 35/24) = 2 . 4
3 - (15/8 + X - 35/24) = 8
15/8 + X - 35/24 = 3 - 8
15/8 + X - 35/24 = -5
15/8 + X = -5 + 35/24
15/8 + X = -85/24
X = -85/24 - 15/8
X = -65/12
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
Mik làm được 1 bài thôi . Tiếc quá mình sắp phải đi học rồi.
BÀi 12:
S=1 + 2 + 22 + `23 +..........+ 22017
2S=2 + 22 + `23 + 24 +..........+22017 + 22018
Trừ đi hai vế ta được:
S=1 + 22018
A= 1 +\(\frac{1}{3}\)+\(\frac{1}{6}\)+ .....+ \(\frac{1}{171}\)+\(\frac{1}{190}\)
A= 1 +2.(\(\frac{1}{6}\)+\(\frac{1}{12}\)+....+\(\frac{1}{342}\)+\(\frac{1}{380}\))
A=1+ 2.(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+....+\(\frac{1}{18.19}\)+\(\frac{1}{19.20}\))
A=1+2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{18}\)-\(\frac{1}{19}\)+\(\frac{1}{19}\)-\(\frac{1}{20}\))
A=1 +2.(\(\frac{1}{2}\)-\(\frac{1}{20}\))
A=1+2.\(\frac{9}{20}\)=1+\(\frac{9}{10}\)=\(\frac{19}{10}\)
B=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+....+\(\frac{1}{2^{20}}\)
2B= 1 +\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+......+\(\frac{1}{2^{21}}\)
2B-B= 1-\(\frac{1}{2^{21}}\)
B=1-\(\frac{1}{2^{21}}\)