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\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).......\left(1+\frac{1}{100}\right)\)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{101}{100}\)
= \(\frac{3.4.5....101}{2.3.4.....100}\)
= \(\frac{101}{2}\)
(1+1/2)(1+1/3)(1+1/4)+...+(1+1/100)
=3/2*4/3*5/4*...*101/100
=101/2
=50,5
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}...\cdot\frac{98}{99}\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
#
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
\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(D=\left(\frac{3}{2\cdot2}\right)\left(\frac{8}{3\cdot3}\right)\left(\frac{15}{4\cdot4}\right)...\left(\frac{9999}{100\cdot100}\right)\)
\(D=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(100\cdot100\right)}\)
\(D=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(D=\frac{1\cdot101}{100\cdot2}\)
\(=\frac{101}{200}\)
\(D=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)(có 50 thừa số nên tích đó là số dương)
\(\Rightarrow D=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\cdot\cdot\cdot\left(\frac{100^2-1}{100^2}\right)\)
\(D=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\cdot\cdot\frac{99\cdot101}{100^2}\)
\(D=\frac{101}{2\cdot100}=\frac{101}{200}\)
VÌ 1/1.1/3.......1/99=2/51.2/52.........2/100
VÀ 2/51.2/52.....2/100=1/1.1/3.......1/99
SUY RA BẰNG NHAU
\(\frac{-1}{7}\left(9\frac{1}{2}-8,75\right)\div\frac{2}{7}+62,5\%\div1\frac{2}{3}\)
\(=\frac{-1}{7}\left(\frac{19}{2}-8,75\right).\frac{7}{2}+62,5\%\div\frac{5}{3}\)
\(=\frac{-1}{7}\left(\frac{19}{2}-\frac{875}{100}\right).\frac{7}{2}+62,5\%.\frac{3}{5}\)
\(=\frac{-1}{2}\left(\frac{38}{4}-\frac{35}{4}\right)+\frac{625}{100}.\frac{3}{5}\)
\(=\frac{-1}{2}.\frac{3}{4}+\frac{25}{4}.\frac{3}{5}\)
\(=\frac{-3}{8}+\frac{75}{20}\)
\(=\frac{-15}{40}+\frac{150}{40}\)
\(=\frac{135}{40}=\frac{27}{8}\)
mình đánh thiếu đề bài ở cuối còn có ''So sánh A với \(-\frac{1}{2}\)