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1/2+1/4+1/8+1/16+1/32
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
=1-1/32
=31/32
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(3B-B=1-\frac{1}{243}\)
\(2B=\frac{242}{243}\)
\(B=\frac{242}{243}\div2\)
\(B=\frac{121}{243}\)
a.A=1/2+1/4+1/8+1/16+1/32+1/64
A= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{4\cdot4}+\frac{1}{4\cdot8}+\frac{1}{8\cdot8}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}\)
= 1 - 1/8 = 7/8
b.B=1/3+1/9+1/27+1/81+1/243
B= \(\frac{1}{1\cdot3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot9}+\frac{1}{9\cdot9}+\frac{1}{9\cdot27}\)
= 1 - 1/27 = 26/27
\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
\(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{16}:0,0625\\ =0,25:0,25-0,125:0,125+0,5:0,5-0,0625:0,0625\\ =1-1+1-1\\ =0+1-1\\ =1-1\\ =0\)
Trả lời:
1/3+1/2+1/5+12/15+22/33+16/32=
(1/3+22/33)+(1/5+12/15)+(1/2+16/32)
=1+1+1=3
A, 32 * 47 + 32 * 53
= 32 * ( 47 + 53 )
= 32 * 100
= 3200 .
B, 875 + 5375 : 25 - 875
= 875 - 875 + 5375 : 25
= 0 + 2115
= 215 .
C, 5 * 25 * 2 * 16 * 4
= ( 5 * 2 * 16 ) * ( 25 * 4 )
= 160 * 100
= 16000 .
1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + ... + 1/256 - 1/512
= 1/2 - 1/512
= 255/512
Gọi \(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\) là A
Ta có :
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(2A=2.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{11}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(A=\frac{1}{2}-\frac{1}{512}\)
\(A=\frac{255}{512}\)
Vậy ..........
đặt A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
A=\(1-\frac{1}{32}\)
A=\(\frac{31}{32}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32
= ( 1/2 + 1/4 ) + ( 1/8 + 1/16 ) + 1/32
= 3/4 + 3/16 + 1/32
= 3/4 + ( 3/16 + 1/32 )
= 3/4 + 7/32
=31/32