Tính nhanh :
A,1/2+1/4+1/8+1/16 +.........+1/256
B,1/3+1/9+1/27+..........+1/729
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a: \(1=4^0\)
\(4=4^1\)
\(16=4^2\)
\(256=4^4\)
b: \(\dfrac{1}{4}=4^{-1}\)
\(\dfrac{1}{64}=4^{-3}\)
\(\dfrac{1}{256}=4^{-4}\)
\(\dfrac{1}{16}=4^{-2}\)
\(\dfrac{1}{1024}=4^{-5}\)
a)
Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:
2/9+6/27+8/36+12/54+16/72+18/81=
2/9+2/9+2/9+2/9+2/9+2/9=
2/9*6=
12/9=
4/3
Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3
b)
Ta có:
1-2/5=3/5
1-2/7=5/7
1-2/9=7/9
...
1-2/99=97/99
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=
3/5*5/7*7/9*...*97/99=
(3*5*7*...*97)/(5*7*9*...*99)=
3/99=
1/33
Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33
c)
Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:
S=1/2+1/4+1/8+1/16+...+1/1024
S*2=1+1/2+1/4+1/8+...+1/512
S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)
S=1-1/1024
S=1023/1024
Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024
a) 1/2 + 3/4 - (3/4 - 4 - 5)
= 1/2 + 3/4 - 3/4 + 4 + 5
= (3/4 - 3/4) + (4 + 5) + 1/2
= 0 + 9 + 1/2
= 19/2
b) [9/16 + 8/(-27)] - (19/27- 7/16 - 2)
= 9/16 - 8/27 - 19/27 + 7/16 + 2
= (9/16 + 7/16) + (-8/27 - 19/27) + 2
= 1 - 1 + 2
= 2
c) -5/8 . [4/9 + 7/(-12)]
= -5/8 . (-5/36)
= 25/288
d) 7/10 . (-3/5) + 7/10 . (-2/5) - (-3/10)
= 7/10 . (-3/5 - 2/5) + 3/10
= 7/10 . (-1) + 3/10
= -2/5
e) -3/7 . 5/9 + 4/9 . (-3/7) + 2 3/7
= -3/7 . (5/9 + 4/9) + 17/7
= -3/7 . 1 + 17/7
= 2
f) 8 2/7 - (3 4/9 + 4 2/7)
= 8 + 2/7 - 3 - 4/9 - 4 - 2/7
= (8 - 3 - 4) + (2/7 - 2/7) - 4/9
= 1 - 4/9
= 5/9
h) 3.(-1/2)² - (4/5 + 8/15) : 5/6
= 3.1/4 - 4/3 : 5/6
= 3/4 - 8/5
= -17/20
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)
A\(\times\) 2 = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)
A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)
A = \(\dfrac{127}{128}\)
Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(B=1+\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+.....+\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(B=1+0-\dfrac{1}{128}\)
\(B=1-\dfrac{1}{128}\)
\(B=\dfrac{128}{128}-\dfrac{1}{128}\)
\(B=\dfrac{127}{128}\)
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)
Bài 1: 1/3+1/9+1/27+1/81+1/243+1/729
Đặt:
A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân A với 3 ta có:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
1/2 + 1/4 + 1/8 +1/16 + 1/32 + 1/64 + 1/128
= 2 . ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 )
= 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 - 1/128 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 ( Rồi giản ước )
= 1