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

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

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

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

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

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

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

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

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

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

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

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

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

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

k cho mk nha

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

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

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

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

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

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

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

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

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

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

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

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 !

Tử số = \(1.2.4+2.3.5+3.4.6+...+100.101.103\)

\(=1.2.\left(3+1\right)+2.3.\left(4+1\right)+3.4.\left(5+1\right)+...+100.101.\left(102+1\right)\)

\(=1.2.3+1.2+2.3.4+2.3+3.4.5+3.4+...+100.101.102+100.101\)

\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)+\left(1.2+2.3+3.4+...+100.101\right)\)

Mẫu số = \(1.2^2+2.3^2+3.4^2+...+100.101^2\)

\(=1.2.\left(3-1\right)+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+100.101.\left(102-1\right)\)

\(=1.2.3-1.2+2.3.4-2.3+3.4.5-3.4+...+100.101.102-100.101\)

\(=\left(1.2.3+2.3.4+3.4.5+...+100.101.102\right)-\left(1.2+2.3+3.4+...+100.101\right)\)

đặt \(A=1.2.3+2.3.4+3.4.5+...+100.101.102\) và \(B=1.2+2.3+3.4+...+100.101\)

bạn tự tính : \(A=\frac{100.101.102.103}{4}=25.101.102.103\); \(B=\frac{100.101.102}{3}=100.101.34\)

rồi thay vào tìm P=\(\frac{A+B}{A-B}\)

\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.....\frac{100^2}{100.101}=\frac{\left(1.2.....100\right).\left(1.2.....100\right)}{\left(1.2.....100\right).\left(2.....101\right)}=\frac{1}{101}\)