\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}...\frac{100.100}{100.101}\)

\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4...100.101}\)

\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3....100\right).\left(2.3.4...101\right)}\)

\(=\frac{1.1}{1.101}\)

\(=\frac{1}{101}\)

\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}.....\frac{100^2}{100\cdot101}\)

\(=\frac{1.1}{1\cdot2}\cdot\frac{2.2}{2.3}\cdot\frac{3.3}{3.4}.....\frac{100.100}{100.101}\)

\(=\frac{\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot100\right)\left(1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot101\right)}\)

\(=\frac{1}{101}\)

(de bai)=\(\frac{1^2.2^2.3^2...100^2}{1.2.2.3.3.4...100.101}\)

=\(\frac{1.1.2.2.3.3.4...100.100}{1.2.2.3.3.4...100.101}\)=\(\frac{1}{101}\)

\(\frac{1.1}{1.2}.\frac{2.2}{2.3}\frac{3.3}{3.4}...\frac{100.100}{100.101}\)

\(=\frac{\left(1.2.3...100\right).\left(1.2.3...100\right)}{\left(1.2.3...100\right).\left(2.3...101\right)}\)

\(=\frac{1}{1.101}\)

\(=\frac{1}{101}\)

k cho mk nha

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.......\frac{99^2}{99.100}.\frac{100^2}{100.101}\)

\(=\frac{1.2.3.....100}{1.2.3....100}.\frac{1.2.3....100}{2.3.4...101}\)

\(=1.\frac{1}{101}=\frac{1}{101}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}\)

\(=\frac{1.2.3...99.100}{2.3.4...100.101}\)

\(=\frac{1}{101}\)

=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)

=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)

tao dóe biet

a,1^2/1.2 . 2^2/2.3 . 3^2/3.4 ... 99^2/99.100 . 100^2/100.101

= 1/2 . 2/3 . 3/4 ... 99/100 . 100/101

=( 2.3.4....100/2.3.4...100) . 1/101

= 1 . 1/101

=1/101

ý b tương tự nhé !

Bạn kia đéo biết thì thôi xen vào làm gì?

tính hộ mình nha

\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)

\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right).\frac{1}{x}=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)

\(\frac{8+4+2+1}{16}.\frac{1}{x}=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(\frac{15}{16}.\frac{1}{x}=3.\left(1-\frac{1}{16}\right)\)

\(\frac{15}{16}.\frac{1}{x}=3.\frac{15}{16}\)

=> \(\frac{1}{x}=3\)

=> \(x=\frac{1}{3}\)

a) = 1-1/2+1/2-1/3+1/3-1/4

    = 1-1/4=3/4

b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018

   =1-1/2018=2017/2018

c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015

   = 1/2-1/2015=2015/4030-2/4030=2013/4030

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)

b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)

\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)

\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}.\frac{2013}{4030}\)

\(=\frac{6039}{8060}\)

]\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)