Tính\(\left(1-\frac{4}{1}\right)\times\left(1-\frac{4}{9}\right)\times\left(1-\frac{1}{25}\right)\times......\times\left(1-\frac{4}{2n-1}\right)\)
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\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{625}\right)\)
\(=\frac{3}{4}\times\frac{8}{9}\times...\times\frac{623}{624}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times...\times\frac{24\times26}{25\times25}\)
\(=\frac{1\times3\times2\times4\times...\times24\times26}{2\times2\times3\times3\times...\times25\times25}\)
Từ đây mình viết nhân là chấm nha mong bạn thông cảm :
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot24\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot26\right)}{\left(2\cdot3\cdot4\cdot...\cdot25\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot25\right)}\)
\(=\frac{1\cdot26}{25\cdot2}\)
\(=\frac{26}{50}=\frac{13}{25}\)
k tớ nha
\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
\(T=(1-\frac{1}{4}).(1-\frac{1}{9}).(1-\frac{1}{16}).....(1-\frac{1}{576}).(1-\frac{1}{625})\)
\(T=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{575}{576}.\frac{624}{625}\)
\(T=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{23.25}{24.24}.\frac{24.26}{25.25}\)
\(T=\frac{1.2.3...23.24}{2.3.4...24.25}.\frac{3.4.5...25.26}{2.3.4...24.25}\)
\(T=\frac{1}{25}.\frac{26}{2}\)
\(T=\frac{1}{25}.13\)
\(T=\frac{13}{25}\)
TK MK NHA
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
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
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)
\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)
\(=\frac{1}{2014}\)
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 :
\(\frac{1000}{2}=500\)
(1+\(\frac{1}{3}\)) x (1+\(\frac{1}{2x4}\)) x(1+\(\frac{1}{3x5}\))x(1+\(\frac{1}{4x6}\)) x .....x (1+ \(\frac{1}{2009x2011}\))
= \(\frac{2}{1x3}\)x \(\frac{2}{2x4}\)x \(\frac{2}{3x5}\)x \(\frac{2}{4x6}\)x....x \(\frac{2}{2009x2011}\)
= ..................
đến đây tự làm nhé
Ta có: \(1-\frac{4}{1}=-3=-\frac{2.1+1}{2.1-1}\)
\(-3.\left(1-\frac{4}{9}\right)=-3.\frac{5}{9}=-\frac{5}{3}=-\frac{2.2+1}{2.2-1}\)
\(-\frac{5}{3}.\left(1-\frac{1}{25}\right)=-\frac{5}{3}.\frac{21}{25}=-\frac{7}{5}=-\frac{2.3+1}{2.3-1}\)
.................................................................................
Vậy kết quả cuối cùng của biểu thức là: \(-\frac{2n+1}{2n-1}\)
Cảm ơn bạn Trần Đình Tuệ