(1/49-1/32).(1/49-1/42).....(1/49-1/492)
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Khách
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LT
2
LT
Lương Thị Vân Anh
CTVHS
27 tháng 1 2023
Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)
= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\)
Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B
Vậy A < B
9 tháng 4 2017
Bài 1:
Ta có:
\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)
\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+...+\left(1+\dfrac{48}{2}\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)
Bài 2:
Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\)
Nhận xét:
\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{2}\)
17 tháng 1 2023
1: =29-16-3*2=13-6=7
2: =35-12+(-14)-2
=23-16=7
3: =4*3125-32:16
=12500-2
=12498
4: =-452+67-75+452=-8
5: =18+99*18-3^5-2^5
=1800-275
=1525
8 tháng 4 2023
= \(\dfrac{49}{2}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}\)
= \(\dfrac{784}{32}-\dfrac{16}{32}-\dfrac{8}{32}-\dfrac{4}{32}-\dfrac{2}{32}-\dfrac{1}{32}\)
= \(\dfrac{753}{32}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4 2023
A = \(\dfrac{49}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\)
A \(\times\) 2 = 49 - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\)
A \(\times\) 2 - A = 48 - \(\dfrac{49}{2}\) + \(\dfrac{1}{32}\)
A = \(\dfrac{1536}{32}\) - \(\dfrac{784}{32}\) + \(\dfrac{1}{32}\)
A = \(\dfrac{753}{32}\)
RM
1
1 tháng 8 2015
(1/49-1/9)(1/49-1/16).....(1/49-1/49)...(1/49-1/2401)
(1/49-1/9)(1/49-1/16).....0....(1/49-1/2401)
=0
vì khoảng cách là 7 mà 49 chia hết cho 7 nên có phân số 1/49 trong dãy
DB
1
21 tháng 8 2017
a, 68 + 42 x 5 - 625 : 25
= 68 + 210 - 25
= 278 - 25
= 253
b, 15 x { 37 + 8 + 20 } + 15 x { 13 + 22 }
= 15 x 65 + 15 x 35
= 15 x { 65 + 35 }
= 15 x 100
= 1500
c, 125 x 56 x 3
= 7000 x 3
= 21000
d, 249 - { 49 - 100 }
= 249 - 49 + 100
= 200 + 100
= 300
e, 1/2 + 1/4 + 1/ 8 + 1/32 + 1/64 + 1/128
= 1/1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128
= 1/1 - 1/128
= 127/128
LT
21 tháng 8 2017
A, \(68+42\times5-625:25\)
= 68 +210 - 25
= 278 - 25
= 253
B,\(15\times\left(37+8+20\right)+15\times\left(13+22\right)\)
= \(15\times\left(37+8+20+13+22\right)\)
=\(15\times100\)
= 1500
C,\(125\times56\times3\)
= \(125\times8\times7\times3\)
= \(1000\times21\)
= 21000
D, 249 - { 49 - 100 }
= 249 - 49 -100
= 200 - 100
= 100
E, \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)(mình sửa lại đề chút xíu nha, nó cứ bị sai sai sao ấy)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
= \(1-\frac{1}{128}\)
=\(\frac{128}{128}-\frac{1}{128}\)
= \(\frac{127}{128}\)