Question

tính bằng cách thuận tiện: dấu ''/ '' là tách nhau thành phân số :)

a.(1 +3+5+7+...+2003+2005)x(125 125x127- 127 127 x125)

b.1 000 x0,04-100:2,5+10x0,4

c.A=1/35+1/63+1/99+...+1/399+1/483

Answer 1

125 125 x 127 - 127 127 x 125

⇒ Dấu cách là dấu gì hả bạn?

Answer 2

a, (1 + 3 + 5 + 7+ ... + 2003 + 2005) x (125125 x 127 - 127127 x125)

= (1 + 3 + 5 +...+2003 + 2005) x (125 x 1001 x 127 - 127x1001 x 125)

= (1 + 3 + 5 +...+ 2003 + 2005) x 0

= 0 

Answer 3

b; 1000 x 0,04 - 100 : 2,5 + 10 x 0,4

= 100 x 0,4 - 100 x 0,4 + 10 x 0,4

= 0 + 4

= 4 

Answer 4

☹

Answer 5

C = \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\) + \(\dfrac{1}{99}\) + ... + \(\dfrac{1}{399}\) + \(\dfrac{1}{438}\)

C = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{2}{5\times7}\) + \(\dfrac{2}{7\times9}\) + \(\dfrac{2}{9\times11}\)+ ...+ \(\dfrac{2}{19\times21}\)) + \(\dfrac{1}{438}\)

C = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{19}\) - \(\dfrac{1}{21}\)) + \(\dfrac{1}{438}\)

C = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{21}\)) + \(\dfrac{1}{438}\)

C = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{16}{105}\) + \(\dfrac{1}{438}\)

C = \(\dfrac{8}{105}\) + \(\dfrac{1}{438}\)

C =  \(\dfrac{401}{5110}\)