Thực hiện các phép tính
B,
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
C,
1/1x2 + 1/2x3 + 1/3x4 + ........ + 1/99x100
D,
( 1-1/2 ) x ( 1- 1/3 ) x ( 1-1/4 ) x ....... x ( 1- 1/100 )
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
XO
3 tháng 7 2019
1) 1/1.2 + 1/2.3 + ... + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7
= 1 - 1/7
= 6/7
2) 1/2 + 1/6 + 1/12 + .. + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
= 1 - 1/9
= 8/9
3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)
= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)
= \(\frac{1.2....2019}{2.3...2020}\)
= \(\frac{1}{2020}\)
4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)
= \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)
A = \(\frac{1}{2}-\frac{1}{2^9}\)
DQ
2
LM
19 tháng 8 2016
A=1-1/2+1/2-1/3+1/3-1/4+...+1/64-1/128
A=1-1/128
A=127/128
Vậy A=\(\frac{127}{128}\)
LM
19 tháng 8 2016
B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
B=1-1/100
B=99/100
Vậy B=\(\frac{99}{100}\)
ND
0
9 tháng 8 2016
a) 3/7 + 4/9 + 4/7 + 5/9
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 7/7 + 9/9
= 1 + 1
= 2
b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5 ) + 5/5
= 2 + 2 + 2 + 2 + 1
= 2 x 4 + 1
= 8 +1
= 9
c) 1/8 + 1/12 + 3/8 + 5/12
= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)
= 4/8 + 6/12
= 1/2 + 1/2
= 2/4 = 1/2
mỏi tay rồi
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 9
d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))
= \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\) x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)
= \(\dfrac{1}{100}\)
PA
8 tháng 7 2016
1.
a.
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
b.
Tích có 100 thừa số
=> n = 100
\(\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times\left(100-100\right)\)
\(=\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times0\)
\(=0\)
2.
a.
\(135\times789789-789\times135135=1001\times\left(135\times789-789\times135\right)=1001\times0=0\)
b.
\(\left(28\times9696-96\times2828\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left[101\times\left(28\times96-96\times28\right)\right]\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=\left(101\times0\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\div\left(1\times2\times3\times...\times2015\times2016\right)\)
\(=0\)
3.
a.
\(\left[\left(x+32\right)-17\right]\times2=42\)
\(\left(x+32\right)-17=\frac{42}{2}\)
\(\left(x+32\right)-17=21\)
\(x+32=21+17\)
\(x+32=38\)
\(x=38-32\)
\(x=6\)
b.
\(125+\left(145-x\right)=175\)
\(145-x=175-125\)
\(145-x=50\)
\(x=145-50\)
\(x=95\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)
\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)
\(< =>\frac{255}{256}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(< =>\frac{1}{1}-\frac{1}{100}\)
\(< =>\frac{99}{100}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(< =>\frac{1}{100}\)
mk chuc ban hoc tot nhe :))