tính giá trị của biểu thức 1/2+1/4+1/8+1/16+1/32+1/64+1/128
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toán 6 nha:
A=1/2+1/4+1/8+1/16+1/32+1/64+1/28
1/128+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128
1/128+A=1/2+1/4+1/8+1/16+1/32+1/64+1/64
1/128+A=1/2+1/4+1/8+1/16+1/32+1/32
1/128+A=1/2+1/4+1/8+1/16+1/16
1/128+A=1/2+1/4+1/8+1/8
1/128+A=1/2+1/4+1/4
1/128+A=1/2+1/2
1/128+A=1
A=1-1/128
a=127/128
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)\(-\frac{1}{64}\)
\(=1-\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}\)\(-\frac{2}{64}-\frac{1}{64}\)
\(=1-\left(\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}-\frac{2}{64}-\frac{1}{64}\right)\)
\(=1-\frac{1}{64}\)
\(=\frac{64}{64}-\frac{1}{64}\)
\(=\frac{63}{64}\)
2A=1+1/2+1/4+1/8+1/16+1/32+1/64
2A-A=(1+1/2+1/4+1/8+1/16+1/32+1/64)-(1/2+1/4+1/8+1/16+1/32+1/64+1/128)
A=1-1/128
A=127/128
A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
suy ra: 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
2A - A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 - 1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128
A = 1 - 1/128 = 127/128
hok tốt
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
Cộng thêm 1/2 vào biểu thức đã cho, có:
S + 1/2= 1/2+1/4+ 1/8+ 1/16+1/32+1/64+1/128
Nhận xét:
127/128
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1x64}{2x64}+\frac{1x32}{4x32}+\frac{1x16}{8x16}+\frac{1x8}{16x8}+\frac{1x4}{32x4}+\frac{1x2}{64x2}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
\(=\left(\frac{64}{128}+\frac{1}{128}\right)+\left(\frac{32}{128}+\frac{8}{128}\right)+\left(\frac{16}{128}+\frac{4}{128}\right)\)
\(=\frac{65}{128}+\frac{40}{128}+\frac{20}{128}\)
\(=125\)