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\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{512}+\frac{1}{1024}=??????????\)
\(< =>1+\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+...+\frac{1}{2\cdot256}+\frac{1}{2\cdot512}\)
\(< =>1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{256}+\frac{1}{2}-\frac{1}{512}\)
\(< =>1+\frac{1}{1}-\frac{1}{512}\)
\(< =>\frac{1023}{512}\)
chuc ban hoc tot nhe :))
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:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)
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 = \(\dfrac{47\times48-47\times47-24-23+2046}{2+4+8+..+512+1024}\)
Đặt tử số là B, mẫu số là C thì
B = 47\(\times\)48 - 47 \(\times\) 47 - 24 -23+2046vàC = 2 + 4 + 8 +....+ 512 + 1024
B = 47 \(\times\) 48 - 47 \(\times\) 47 - 24 - 23 + 2046
B= 47 \(\times\) 48 - 47 \(\times\) 47 - ( 24 + 23) + 2046
B = 47 \(\times\) 48 - 47 \(\times\) 47 - 47 + 2046
B = 47 \(\times\) 48 - 47 \(\times\) 47 - 47 \(\times\) 1 + 2046
B = 47 \(\times\) ( 48 - 47 - 1) + 2046
B = 47 \(\times\) 0 + 2046
B = 2046
C = 2 + 4 + 8+ ....+ 512 +1024
C \(\times\) 2 = 4 + 8 +.....+ 512 + 1024 + 2048
C \(\times\) 2 - C = 2048 - 2
C \(\times\) ( 2 - 1) = 2046
C = 2046
A = \(\dfrac{B}{C}\) = \(\dfrac{2046}{2046}\) = 1
a)
\(A=2+4+8+...+2048\)
\(A=2+2^2+...+2^{11}\)
\(2A=2^2+2^3+...+2^{12}\)
\(2A-A=\left(2^2+2^3+...+2^{12}\right)-\left(2+2^2+...+2^{11}\right)\)
\(A=2^{12}-2\)
a) 2+4+8+16+32+64+128+512+1024+2048
b thì minh cha ra. Mình sẽ cố làm ra b mong ban thong cam va bạn nho k đung cho minh nha
A = 1/2 + 1/4 + 1/8 + ... + 1/1024
2A = 1 + 1/2 + 1/4 + ... + 1/512
2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)
A = 1 - 1/1024
A = 1023/1024
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)
\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
\(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}+\frac{3}{256}+\frac{3}{512}+\frac{3}{1024}\)
=\(3.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\right)\)
=\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\right)\)
=\(3.\left(1-\frac{1}{1024}\right)=3.\left(\frac{1024}{1024}-\frac{1}{1024}\right)=3.\frac{1023}{1024}=\frac{3069}{1024}\)
Chúc em học tốt
2046-(47.48-47.47-20-27)
=2046-(47-20-27)
=2046
2+4+6+8+16+32+.....+512+1024
bài còn lại mình đang suy nghĩ nhé
bao giờ ra mình gửi cho
Đừng vì thế mà ko đánh k cho mình nha