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a) Đặt A= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2A= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
= 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=>A = 2A-A =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 - 1/256
=1-1/256
=255/256
b) = 18 - (19/21 + 8/9) = 18 - 113/63 = 1021/63
1.a
1/2+1/4+1/8+1/16+1/32
= 1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
= 1-1/32=31/32
1b
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3} +\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{2}{3}+\frac{1}{4}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{5}{20}+\frac{5}{30}+\frac{20}{30}+\frac{5}{20}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{5}{20}+\frac{5}{20}+\frac{1}{20}\right)+\left(\frac{5}{30}+\frac{20}{30}+\frac{1}{30}\right)\)
\(=\frac{11}{20}+\frac{26}{30}\)
\(=\frac{11}{20}+\frac{13}{15}\)
\(=\frac{17}{12}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
=> \(A=1-\frac{1}{32}=\frac{31}{32}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
\(=\frac{63}{64}\)
Ta có:\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)\(\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}\)\(+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\)\(\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)\(=\frac{63}{64}\)
a) 1/2+1/4+1/8+1/16+1/32
Đặt A = 1/2+1/4+1/8+1/16+1/32
A x 2 = 2 x (1/2+1/4+1/8+1/16+1/32 )
= 1 + 1/2 + 1/4 +1/6
Lấy A x 2 - A ta có :
A x 2 - A = 1 + 1/2+1/4+1/8+1/16+ - 1/2+1/4+1/8+1/16+1/32
A = 1 - 1/32
A = 31/32
Câu b dễ nên tự làm đi
Cách 1:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64
B=1-1/64
B=63/64
Cách 2:
B=1/2+1/4+1/8+1/16+1/32+1/64
B=1/21+1/22+1/23+1/24+1/25+1/26
2B=1+1/21+1/2^2+1/2^3+1/2^4+1/2^5
2B-B=1-1/2^6
B=1-1/64
B=63/64
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)
\(=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)
a) \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\)
\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)
1+1=2
2+2=4
4+4=8
8+8=16
16+16=32
32+32=64
64+64=128
k nhẽ mk nhanh nhất
\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)
\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)
Thay vào ta có :
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
...................
\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
Thay vào ta có :
\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)
\(=1-\frac{1}{256}\)
\(=\frac{255}{256}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)
=\(\frac{16+8+4+2+1}{32}=\frac{31}{32}\)