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a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
a x 3 - a = (1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243) - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
a x 2 = 1 - 1/729
a x 2 = 728/729
a = 728/729 : 2 = 364/729
Với câu a : \(\frac{1}{2}+\frac{1}{3}\times\frac{1}{6}=\frac{1}{2}+\frac{1}{18}=\frac{9}{18}+\frac{1}{18}=\frac{10}{18}=\frac{5}{9}\)
Ta không chọn câu a vì \(\frac{5}{9}\ne\frac{5}{36}\)khi quy đồng lên thành \(\frac{20}{36}\ne\frac{5}{36}\)
Với câu b : \(\frac{1}{9}+\frac{5}{12}-\frac{7}{18}=\frac{4}{36}+\frac{15}{36}-\frac{14}{36}=\frac{4+15-14}{36}=\frac{5}{36}\)
Ta chọn câu b vì kết quả bằng nhau theo đề bài trên
Còn các câu còn lại đều bị loại
a) đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow2\times A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Rightarrow2\times A-A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
phần b bn cx lm tương tự như z nha!
c) sửa đề:
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{13x14}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)
\(=1-\frac{1}{14}=\frac{13}{14}\)
Sửa:
d) \(\frac{1}{15x18}+\frac{1}{18x21}+\frac{1}{21x24}+...+\frac{1}{87x90}\)
\(=\frac{1}{3}x\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(=\frac{1}{3}x\frac{1}{18}\)
\(=\frac{1}{54}\)
\(A=\dfrac{3}{2}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}\)
\(A=\dfrac{3}{2}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\right)\)
Đặt Y = \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
=>\(2Y=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\)
\(2Y-Y=1-\dfrac{1}{256}\)
\(Y=\dfrac{255}{256}\)
=> A = \(\dfrac{3}{2}-\dfrac{255}{256}=\dfrac{129}{256}\)
B = \(2-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-...-\dfrac{1}{729}\)
\(B=2-\left(\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\right)\)
Đặt E=\(\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\)
\(\Rightarrow3E=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+..+\dfrac{1}{243}\)
\(3E-E=\dfrac{1}{3}-\dfrac{1}{729}\)
\(2E=\dfrac{242}{729}\)
\(E=\dfrac{121}{729}\)
=> B = \(2-\dfrac{121}{729}=\dfrac{1337}{729}\)
C = \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{99}\)
\(C=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{9\cdot11}\)
C = \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{9\cdot11}\right)\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(C=\dfrac{1}{2}\left(1-\dfrac{1}{11}\right)\)
\(C=\dfrac{1}{2}\cdot\dfrac{10}{11}=\dfrac{5}{11}\)
D = \(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{19\cdot24}+..+\) \(\dfrac{1}{1999\cdot2004}\)
D = \(\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{1999\cdot2004}\right)\)
\(D=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)
\(D=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)\)
\(D=\dfrac{1}{5}\cdot\dfrac{125}{501}\)
\(D=\dfrac{25}{501}\)
có ai giải hộ ko?