a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

=  1 + 1 + 8

=  2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)

=  \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8

= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8

= \(\dfrac{1}{2}\) + 4

= \(\dfrac{9}{2}\) 

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

  = (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

= 1 + 1 + 8

= 2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)

= \(\dfrac{1}{2}\) x 10

= 5

Gọi tổng là A ta có:

\(A.2=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{18.20}\)

\(A.2=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\)

\(A.2=\frac{1}{2}-\frac{1}{20}\)

\(A=\frac{9}{20}:2=\frac{9}{40}\)

1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+9+8+9+10+12+13+14+15+16+17+18+19+20+30+1000000=10000274

10000274

a:

Số số hạng của dãy số 1;2;3;...;102 là:

\(\frac{102-1}{1}+1=\frac{101}{1}+1=101+1=102\) (số)

1-2+3-4+5-6+...+101-102+103

=(1-2)+(3-4)+(5-6)+...+(101-102)+103

\(=\left(-1\right)+\left(-1\right)+\cdots+\left(-1\right)+103\)

\(=\left(-1\right)\cdot\frac{102}{2}+103=103-51=52\)

b: Sửa đề: 2-4+6-8+10-12+...+98-100+102

Số số hạng của dãy số 2;4;6;8;...;98;100 là:

\(\frac{100-2}{2}+1=\frac{98}{2}+1=49+1=50\) (số)

2-4+6-8+10-12+...+98-100+102

=(2-4)+(6-8)+(10-12)+...+(98-100)+102

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{100}{2}+102\)

=-100+102

=2

c: Số số hạng trong dãy 16;18;20;22;...;64;66 là:

\(\frac{66-16}{2}+1=\frac{50}{2}+1=25+1=26\) (số)

16-18+20-22+...+64-66+68

=(16-18)+(20-22)+...+(64-68)+68

=(-2)+(-2)+...+(-2)+68

\(=\left(-2\right)\cdot\frac{26}{2}+68\)

=68-13

=55

rễ ,chỉ cần gạch thôi

dễ lấy máy tính ra bấm là 618

mình xin nhầm đề là:   A = \(1*2*3+2*4*6+4*8*12+8*16*24 \over2*3*4+4*6*8+8*12*16+16*24*32\)