Cho A= 26/25 + 37/36 + 50/49 +...+ 10001/10000 . Chứng minh rằng: 96/1/6 < A < 96/1/4
Tính nhanh.
a) (-4).(-8).(-17).(-25).125
b) (236 – 362) – (638 + 463)
c) (- 325 – 147) – ( - 625 + 853) – (-96)
d) 69.(- 47) – 31. (- 47)
e) 98. (- 37) + (- 49).126
f) – 35. (49 – 125) – 125 .(35 – 49)
g) 2 + 4 – 6 – 8 + 10 + 12 – 14 – 16 + …+ 90 + 92 – 94 – 96
h) 1 + 3 + 3 2 + 3 3 + ….+ 3 50
A) Tính M: 3/4.8/9.15/16.9999/10000 B) Chứng tỏ rằng: 1/26+1/27+...+1/50=99/50-97/49+...+7/4-5/3+3/2-1
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)
\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
Xét vế phải :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
Tính hợp lí
a, 15.(27+ 18 +6 ) + 15.( 23 +12)
b, 24.(15 + 49 ) + 12. ( 50 + 42 )
c, 53.(51+4)+53.(49+96) +53
d, 42.( 15 +96) +6.(25 +4).7
Tính hợp lí
a,15.(27+ 18 +6 ) + 15.( 23 +12)
= 15(27+18+6+23+12)
= 15.86
= 1290
b,24.(15 + 49 ) + 12. ( 50 + 42 )
=12.2(15+49)+12.(50+42)
= 12(30+98)+12(50+42)
=12(30+98+50+42)
=12.220
= 2640
c,53.(51+4)+53.(49+96) +53
=53(51+4)+53(49+96)+53.1
=53(51+4+49+96+1)
=53.201
=10653
d,42.( 15 +96) +6.(25 +4).7
=42(15+96)+42(25+4)
=42(15+96+25+4)
=42.140
=5880
tick đúng cho mk nha pikachu
Tính hợp lí:
\(a)\)\(15.\left(27+18+6\right)+15.\left(23+12\right)\)
\(=\)\(15.51+15.35\)
\(=\)\(15.\left(51+35\right)\)
\(=\)\(15.86\)
\(=\)\(1290\)
\(b)\)\(24.\left(15+49\right)+12.\left(50+42\right)\)
\(=\)\(24.64+12.92\)
\(=\)\(1536+1104\)
\(=\)\(2640\)
\(c)\)\(53.\left(51+4\right)+53.\left(49+96\right)+53\)
\(=\)\(53.55+53.145+53\)
\(=\)\(53.\left(145+55+1\right)\)
\(=\)\(53.201\)
\(=\)\(10653\)
\(d)\)\(42.\left(15+96\right)+6.\left(25+4\right).7\)
\(=\)\(42.\left(15+96\right)+42.\left(25+4\right)\)
\(=\)\(42.111+42.29\)
\(=\)\(42.140\)
\(=\)\(5880\)
cho :
A = 1 - 1/2 + 1/3 - 1/4 + ...+ 1/49 - 1/50 ; B = 1/25 + 1/26 + 1/27 +...+1/50
chứng minh A = B
1. So sanh
a) 42/43 va 58/59
b) 18/31 va 15/37
c) 53/57 va 531/571
d) 25/26 và 25251/26261
e) 3535.232323/353535.2323; 3535/3534 va 2323/2322
g) 13/38 va 1/3
2.Chung minh rang:
a) N=1/4^2+1/6^2+1/8^2+...+1/(2n)^2 < 1/4
b) 1/26+1/27+1/28+...+1/50=1-1/2+1/3-1/4+...+1/49-1/50
c) A=1/6+1/24+1/60+...+1/6840 < 1/4
d) B= 36/15+36/105+36/315+...+36/19575 < 3
Mn oi!!!!! Giup minh lam 2 bai nay voi!!!!!
Bài 1 :
a) \(\dfrac{42}{43}=1-\dfrac{1}{43}\)
\(\dfrac{58}{59}=1-\dfrac{1}{59}\)
Mà \(\dfrac{1}{43}>\dfrac{1}{59}\Leftrightarrow\dfrac{42}{43}< \dfrac{58}{59}\)
b) \(\dfrac{18}{31}>\dfrac{15}{31}>\dfrac{15}{37}\)
\(\Leftrightarrow\dfrac{18}{31}>\dfrac{15}{37}\)
c) \(\dfrac{53}{57}=1-\dfrac{4}{57}\)
\(\dfrac{531}{517}=1-\dfrac{40}{517}\)
Mà \(\dfrac{4}{57}=\dfrac{40}{570}>\dfrac{40}{517}\)
\(\Leftrightarrow\dfrac{53}{57}< \dfrac{531}{517}\)
Chứng minh rằng:
1/1×2 + 1/3×4 + 1/5×6 +.......+1/49×50=1/26 + 1/27 + 1/28 +.....+1/50
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
Tính nhanh
13.58.4 + 32.26.2+52.10
15.34.4+120.21+21.5.12
14.35.5+10.25.7+20.70
24.( 15+49)+12.(50+42)
53.(51+4)+53.(49+96)+53
42.(15+96)+6.(25+4).7
45.(13+78)+8.(53+25).2
cọu lên gg í <<:
Bài 1: Chứng minh rằng:
a) 165+ 215 chia hết cho 33
b) 88+ 220 chia hết cho 17
c) 4343 - 1717 chia hết cho 10
d) 1 - 2 + 22 - 23 + 24 - 25 + 26 - ... - 22021 + 22022 chia 6 dư 1
Bài 2: Chứng minh rằng:
a) \(\overline{aaa}\) ⋮ 37 b) (\(\overline{ab}\) + \(\overline{ba}\)) ⋮ 11
Bài 1
a, cm : A = 165 + 215 ⋮ 3
A = 165 + 215
A = (24)5 + 215
A = 220 + 215
A = 215.(25 + 1)
A = 215. 33 ⋮ 3 (đpcm)
b,cm : B = 88 + 220 ⋮ 17
B = (23)8 + 220
B = 216 + 220
B = 216.(1 + 24)
B = 216. 17 ⋮ 17 (đpcm)
c, cm: C = 1 - 2 + 22 - 23 + 24 - 25 + 26 -...-22021 + 22022 : 6 dư 1
C=1+(-2+22-23+24- 25+26)+...+(-22017+22018-22019+22020-22021+22022)
C = 1 + 42 +...+ 22016.(-2 + 22 - 23 + 24 - 25 + 26)
C = 1 + 42+...+ 22016.42
C = 1 + 42.(20+...+22016)
42 ⋮ 6 ⇒ C = 1 + 42.(20+...+22016) : 6 dư 1 đpcm
a, \(\overline{aaa}\) \(⋮\) 37
\(\overline{aaa}\) = a x 111 = a x 3 x 37 ⋮ 37 (đpcm)
b, (\(\overline{ab}\) + \(\overline{ba}\)) ⋮ 11
\(\overline{ab}\) + \(\overline{ba}\) = \(\overline{a0}\) + b + \(\overline{b0}\) + a = \(\overline{aa}\) + \(\overline{bb}\) = a x 11 + b x 11 = 11 x (a+b)⋮11
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