a)   \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)

\(=\frac{1}{5}\)

b)   \(\left(1-\frac{3}{4}\right).\left(1-\frac{3}{7}\right).\left(1-\frac{3}{10}\right)........\left(1-\frac{3}{97}\right).\left(1-\frac{3}{100}\right)\)

\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.......\frac{94}{97}.\frac{97}{100}\)

\(=\frac{1}{100}\)

\(D=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)

\(3D=\frac{2.3}{1.4}+\frac{2.3}{4.7}+...+\frac{2.3}{97.100}\)

\(3D=2\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(3D=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(3D=2\left(1-\frac{1}{100}\right)\)

\(3D=2\cdot\frac{99}{100}\)

\(3D=\frac{99}{50}\)

\(D=\frac{99}{50}:3\)

\(D=\frac{33}{50}\)

B = \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

= \(\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

a) B= 1 + 3 + 7 + 9 +...+ 99

= ( 1 + 99 ) + ( 97 + 3 ) + ( 95 + 5 ) + ( 93 + 7 ) + ( 91 + 9 ) + ( 89 + 11 ) + ... ,

(Tổng cộng có 25 cặp có tổng là 100.)
= 25 x 100.
B = 2500.

b,    C = 1+4+7+10+..............+100

Số số hạng là : 

(100-1) : 3 + 1 = 34 (số)

C = (100+1) x 34 : 2 

C = 101 x 34  : 2

C = 3434 : 2

C = 1717

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(A=\frac{3^2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\cdot\left(1-\frac{1}{100}\right)\)

\(A=3\cdot\frac{99}{100}=\frac{297}{100}\)

Vậy \(A=\frac{297}{100}\)

\(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+\cdot\cdot\cdot+\dfrac{2}{97\times100}\)

\(=\dfrac{2}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\cdot\cdot\cdot+\dfrac{3}{97\times100}\right)\)

\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\cdot\cdot\cdot+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{2}{3}\times\dfrac{99}{100}\)

\(=\dfrac{33}{50}\)

#\(Toru\)

Công thức: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\left(n\ne0;n\ne-a\right)\)

Ta đặt biểu thức là :

A = 2/1 x 4 + 2/4 x 7  + 2/7 x 10 + ... + 2/97 x 100

A = 2 - 2/4 + 2/4 - 2/7 + 2/7 - 2/10 + ... + 2/97 - 2/100

A = 2 - 2 /100

A = 99/50

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

= \(\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{2}{3}\left(1-\frac{1}{100}\right)\)

= \(\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

A = 1 + 3 + 5 + ... + 101

A = ( 101 + 1) x 51 : 2

A = 2061

B = 1 + 4 + 7 + 10 + ...+ 100

B = ( 1 + 100) x 34 :2

B = 1717

E=1+2-3-4+5+6-7-8+...-299-300+301+302

=1+(2-3-4)+5+(6-7-8)+...+(298-299-300)+301+302

=1-5+5-9+...-301+301+302

=1+302=303