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\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
biết làm bài 1 thôi
\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)
= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)
lượt bỏ đi còn :
\(\frac{1000}{2}=500\)
A = 1/2 x 2/3 x.....x 2009/2010 = 1x2x3x....x2009 / 2x3x4x.........x 2010
A = 1/ 2010
vậy A = 1/ 2010
\(A=\frac{1}{2}\times\frac{2}{3}\times...\times\frac{2009}{2010}\)
\(A=\frac{1\times2\times...\times2009}{2\times3\times2010}\)
\(A=\frac{1}{2010}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\cdot\cdot\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}\)
\(=\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{100}{2}\)
\(=50\)
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+...+16\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{16}.16.17:2=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}=\frac{2+3+4+...+17}{2}=\frac{152}{2}=76\)
\(A=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot...\cdot\frac{1325}{1326}\)
\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot...\cdot\frac{2650}{2652}\)
\(=\frac{5\cdot8}{6\cdot7}\cdot\frac{6\cdot9}{7\cdot8}\cdot\frac{7\cdot10}{8\cdot9}\cdot...\cdot\frac{50\cdot53}{51\cdot52}\)
\(=\frac{5\cdot53}{7\cdot51}=\frac{265}{357}\)
Từ đề bài suy ra: A=3/4 x 8/9 x ...x 9800/9801 x 9999/10000
=>A=<1x3/2x2> x <2x4/3x3> x ... x <99x101/100x100>
=>A=(1x2x...x99)/(2x3x...x100) x (3x4x...x101)/(2x3x...x100)
=>A=1/100 x 101/2 = 101/200
BẠn hãy tính ra rồi phân tích tử và mẫu là ra