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Ta có: \(\frac{1}{2}=1-\frac{1}{2}\); \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\); \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\); ...; \(\frac{1}{512}=\frac{1}{256}-\frac{1}{512}\); \(\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
Vậy \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)
\(=1+1-\frac{1}{1024}\)
\(=2-\frac{1}{1024}=\frac{2047}{1024}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
quy đồngcác phân số lấy mẫu số là 512 .ta có tử số là
256 +128 + 64 +32 +16 +8 +4 +2 +1 =495
A =\(\frac{495}{512}\)
A=1999/2000
B=199/200
C=511/512
hok tốt
Đáp án
mình lười trình bày cách làm lém, để đáp án thui nha
A = \(\frac{1999}{2000}\)
B = \(\frac{199}{200}\)
C = \(\frac{511}{512}\)
= 1/2+1/4+....+1/512+1/512 - 1/512
= 1/2+1/4+....+1/256+1/256 - 1/512
........
= 1/2+1/2 - 1/512 = 1-1/512 = 511/512
k mk nha
1)
a) \(x+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=5\)
\(x+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}=5\)
\(x+\frac{127}{128}=5\)
\(x=5-\frac{127}{128}=\frac{513}{128}\)
b) \(x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}=3\)
\(x+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}=3\)
\(x+\frac{2186}{2187}=3\)
\(x=3-\frac{2186}{2187}=\frac{4375}{2187}\)
2)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
b) \(5\frac{1}{2}+3\frac{5}{6}+\frac{2}{3}\)
\(=\left(5+3\right)+\left(\frac{1}{2}+\frac{2}{3}+\frac{5}{6}\right)\)
\(=8+\left(\frac{3}{6}+\frac{4}{6}+\frac{5}{6}\right)\)
\(=8+2=10\)
c) \(7\frac{7}{8}+1\frac{4}{6}+3\frac{3}{5}\)
\(=\left(7+1+3\right)+\left(\frac{7}{8}+\frac{2}{3}+\frac{3}{5}\right)\)
\(=11+\left(\frac{105}{120}+\frac{80}{120}+\frac{72}{120}\right)\)
\(=11+\frac{257}{120}=\frac{1577}{120}\)
3) Gọi số đó là x. Theo đề ta có :
\(\frac{16-x}{21+x}=\frac{5}{7}\)
\(7\left(16-x\right)=5\left(21+x\right)\)
\(112-7x=105+5x\)
\(112-105=7x-5x\)
\(7=2x\)
\(x=\frac{7}{2}=3,5\) ( vô lí )
Vậy không có số tự nhiên để thõa mãn điều kiện trên.
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
ta có: 1+1/2+2+1/4+...+9+1/512
=(1+2+3+4+...+9)+(1/2+1/4+...+1/512)
=45+(1/2+1/4+...+1/512)
gọi số hạng (1/2+1/4+...+1/512) là a ta được :
a=1/2+1/4+...+1/512
2a=1+1/2+1/4+1/8+...+1/256
2a-a=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)
=1-1/512
=511/512
vậy kết quả của biểu thức đó là45+511/512